Why did C use the -> operator instead of reusing the . operator?












17















In the C programming language, the syntax to access the member of a structure is




structure.member




However, a member of a structure referenced by a pointer is written as




pointer->member




There's really no need for two different operators. The compiler knows the type of the left-hand value; if it is a structure, the first meaning is evident. If it is a pointer, the second meaning is evident. Furthermore, . is far easier to type than ->. Not only does -> have more characters to type, on many keyboards one character is unshifted and the other character is shifted, requiring some finger acrobatics. Indeed, many languages based on C allow or use . in place of ->.



Why did C use two operators when one would have sufficed?



(My guess would be because C evolved from the typeless B language.)










share|improve this question


















  • 3





    You can still write (*structure).member, if you like that more. (I don't, very probably K&R didn't, either. It is a bit awkward to handle because of C operator precedence, and that might answer your question)

    – tofro
    7 hours ago













  • @tofro: True, such a form always was possible, and avoids introducing another operator. However, it is far worse in terms of finger acrobatics and (as BrianH points out) readability.

    – Dr Sheldon
    4 hours ago






  • 4





    Just out of curiosity: In your hypothetical language where a.b could be interpreted as (*a).b if a is a pointer-to-struct, would it also be automatically be interpreted as (**a).b if a is a pointer-to-pointer-to-struct? Just to point out a possible consequence…

    – wrtlprnft
    4 hours ago











  • @wrtlprnft: Interesting thought. I can see arguments both for and against such behavior, so I'm not sure there is a clear answer.

    – Dr Sheldon
    4 hours ago






  • 2





    People complain a lot about pointers being confusing. Imagine adding to that confusion by not knowing if a variable was a pointer or not when reading code.

    – JPhi1618
    4 hours ago
















17















In the C programming language, the syntax to access the member of a structure is




structure.member




However, a member of a structure referenced by a pointer is written as




pointer->member




There's really no need for two different operators. The compiler knows the type of the left-hand value; if it is a structure, the first meaning is evident. If it is a pointer, the second meaning is evident. Furthermore, . is far easier to type than ->. Not only does -> have more characters to type, on many keyboards one character is unshifted and the other character is shifted, requiring some finger acrobatics. Indeed, many languages based on C allow or use . in place of ->.



Why did C use two operators when one would have sufficed?



(My guess would be because C evolved from the typeless B language.)










share|improve this question


















  • 3





    You can still write (*structure).member, if you like that more. (I don't, very probably K&R didn't, either. It is a bit awkward to handle because of C operator precedence, and that might answer your question)

    – tofro
    7 hours ago













  • @tofro: True, such a form always was possible, and avoids introducing another operator. However, it is far worse in terms of finger acrobatics and (as BrianH points out) readability.

    – Dr Sheldon
    4 hours ago






  • 4





    Just out of curiosity: In your hypothetical language where a.b could be interpreted as (*a).b if a is a pointer-to-struct, would it also be automatically be interpreted as (**a).b if a is a pointer-to-pointer-to-struct? Just to point out a possible consequence…

    – wrtlprnft
    4 hours ago











  • @wrtlprnft: Interesting thought. I can see arguments both for and against such behavior, so I'm not sure there is a clear answer.

    – Dr Sheldon
    4 hours ago






  • 2





    People complain a lot about pointers being confusing. Imagine adding to that confusion by not knowing if a variable was a pointer or not when reading code.

    – JPhi1618
    4 hours ago














17












17








17


1






In the C programming language, the syntax to access the member of a structure is




structure.member




However, a member of a structure referenced by a pointer is written as




pointer->member




There's really no need for two different operators. The compiler knows the type of the left-hand value; if it is a structure, the first meaning is evident. If it is a pointer, the second meaning is evident. Furthermore, . is far easier to type than ->. Not only does -> have more characters to type, on many keyboards one character is unshifted and the other character is shifted, requiring some finger acrobatics. Indeed, many languages based on C allow or use . in place of ->.



Why did C use two operators when one would have sufficed?



(My guess would be because C evolved from the typeless B language.)










share|improve this question














In the C programming language, the syntax to access the member of a structure is




structure.member




However, a member of a structure referenced by a pointer is written as




pointer->member




There's really no need for two different operators. The compiler knows the type of the left-hand value; if it is a structure, the first meaning is evident. If it is a pointer, the second meaning is evident. Furthermore, . is far easier to type than ->. Not only does -> have more characters to type, on many keyboards one character is unshifted and the other character is shifted, requiring some finger acrobatics. Indeed, many languages based on C allow or use . in place of ->.



Why did C use two operators when one would have sufficed?



(My guess would be because C evolved from the typeless B language.)







c






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









Dr SheldonDr Sheldon

1,9762834




1,9762834








  • 3





    You can still write (*structure).member, if you like that more. (I don't, very probably K&R didn't, either. It is a bit awkward to handle because of C operator precedence, and that might answer your question)

    – tofro
    7 hours ago













  • @tofro: True, such a form always was possible, and avoids introducing another operator. However, it is far worse in terms of finger acrobatics and (as BrianH points out) readability.

    – Dr Sheldon
    4 hours ago






  • 4





    Just out of curiosity: In your hypothetical language where a.b could be interpreted as (*a).b if a is a pointer-to-struct, would it also be automatically be interpreted as (**a).b if a is a pointer-to-pointer-to-struct? Just to point out a possible consequence…

    – wrtlprnft
    4 hours ago











  • @wrtlprnft: Interesting thought. I can see arguments both for and against such behavior, so I'm not sure there is a clear answer.

    – Dr Sheldon
    4 hours ago






  • 2





    People complain a lot about pointers being confusing. Imagine adding to that confusion by not knowing if a variable was a pointer or not when reading code.

    – JPhi1618
    4 hours ago














  • 3





    You can still write (*structure).member, if you like that more. (I don't, very probably K&R didn't, either. It is a bit awkward to handle because of C operator precedence, and that might answer your question)

    – tofro
    7 hours ago













  • @tofro: True, such a form always was possible, and avoids introducing another operator. However, it is far worse in terms of finger acrobatics and (as BrianH points out) readability.

    – Dr Sheldon
    4 hours ago






  • 4





    Just out of curiosity: In your hypothetical language where a.b could be interpreted as (*a).b if a is a pointer-to-struct, would it also be automatically be interpreted as (**a).b if a is a pointer-to-pointer-to-struct? Just to point out a possible consequence…

    – wrtlprnft
    4 hours ago











  • @wrtlprnft: Interesting thought. I can see arguments both for and against such behavior, so I'm not sure there is a clear answer.

    – Dr Sheldon
    4 hours ago






  • 2





    People complain a lot about pointers being confusing. Imagine adding to that confusion by not knowing if a variable was a pointer or not when reading code.

    – JPhi1618
    4 hours ago








3




3





You can still write (*structure).member, if you like that more. (I don't, very probably K&R didn't, either. It is a bit awkward to handle because of C operator precedence, and that might answer your question)

– tofro
7 hours ago







You can still write (*structure).member, if you like that more. (I don't, very probably K&R didn't, either. It is a bit awkward to handle because of C operator precedence, and that might answer your question)

– tofro
7 hours ago















@tofro: True, such a form always was possible, and avoids introducing another operator. However, it is far worse in terms of finger acrobatics and (as BrianH points out) readability.

– Dr Sheldon
4 hours ago





@tofro: True, such a form always was possible, and avoids introducing another operator. However, it is far worse in terms of finger acrobatics and (as BrianH points out) readability.

– Dr Sheldon
4 hours ago




4




4





Just out of curiosity: In your hypothetical language where a.b could be interpreted as (*a).b if a is a pointer-to-struct, would it also be automatically be interpreted as (**a).b if a is a pointer-to-pointer-to-struct? Just to point out a possible consequence…

– wrtlprnft
4 hours ago





Just out of curiosity: In your hypothetical language where a.b could be interpreted as (*a).b if a is a pointer-to-struct, would it also be automatically be interpreted as (**a).b if a is a pointer-to-pointer-to-struct? Just to point out a possible consequence…

– wrtlprnft
4 hours ago













@wrtlprnft: Interesting thought. I can see arguments both for and against such behavior, so I'm not sure there is a clear answer.

– Dr Sheldon
4 hours ago





@wrtlprnft: Interesting thought. I can see arguments both for and against such behavior, so I'm not sure there is a clear answer.

– Dr Sheldon
4 hours ago




2




2





People complain a lot about pointers being confusing. Imagine adding to that confusion by not knowing if a variable was a pointer or not when reading code.

– JPhi1618
4 hours ago





People complain a lot about pointers being confusing. Imagine adding to that confusion by not knowing if a variable was a pointer or not when reading code.

– JPhi1618
4 hours ago










3 Answers
3






active

oldest

votes


















19














In the embryonic form of C described in the 1974 C Reference Manual, there was no requirement that the left operand of . actually be a structure, nor that the left operand of -> actually be a pointer. The -> operator meant "interpret the value of the left operand as a pointer, add the offset associated with the indicated structure member name, and dereference the resulting pointer as an object of the appropriate type. The . operator effectively took the address of the left operand and then applied ->.



Thus, given:



struct q { int x, y; };
int a[2];


the expressions a[0].y and a[0]->y would be interpreted in a fashion equivalent to ((struct q*)&a[0])->y and ((struct q*)a[0])->y, respectively.



If the compiler had examined the type of the left operand to the . operator, it could have used that to select between the two behaviors for it. It was probably easier, however, to have two operators whose behaviors didn't depend upon the left operand's type.






share|improve this answer



















  • 2





    To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

    – Raffzahn
    5 hours ago



















12














I think there are two factors that led to standardization of the distinct operator "->" for accessing data members using a pointer.




  1. You assume that the C compiler would recognize the type of the LHS as being a pointer. But programmers could, and often did, override the initial typing (variable declaration) by using a typecast.

  2. In order to make the code more readable and less prone to unintended side-effects, it is useful to distinguish operations using pointers.


A very common feature of idiomatic C code is that a structure passed to a function as a pointer is modified within the function. Thus, the result is returned implicitly, by the side-effect of the structure variable in the calling function having been modified by the callee. This sort of approach violates modern sensibilities about loosely coupled code, but it was a simple and efficient means of dealing with complexity in C code. I would say the programmer was greatly assisted in maintaining the readability of such code by having distinct operations that made it clear whether some (possibly shared) memory pointer was the thing whose target was being modified.






share|improve this answer


























  • Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

    – Dr Sheldon
    7 hours ago








  • 1





    @DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

    – Brian H
    6 hours ago





















9














Despite your assertion, there would in fact be situations where it would be ambigious.



First off, early C compilers were very simple. This was in fact the main appeal of the language, as compilers for it were very easy to create and could run on very small systems, like early 16/32 microprocessors.



Adding a bunch of code for hitting all the niche cases of type inference would have drastically added to the amount of code required to make a C compiler. In fact, I've argued as a (half) joke that K&R C had type inference, but it always inferred int. If you didn't tell C what type an object was, it assumed int (which could cause some really gnarly bugs, let me tell you...)



Secondly, since K&R C was weakly(barely) typed, the information in many cases flat out wasn't available. The destination type of a pointer assignment can be an int, or visa versa, and K&R C has no problem with that. The compiler simply cannot infer a dereference. The coder is assumed to know what she's doing.



Also realize that in C pointers and arrays are essentially syntactic sugar for each other. This means now your . operator would have to automagically work on arrays too. For instance, if member happened to be a char array, now structure.member would return with the first character. And again, both chars and pointer are assignable into ints, so context doesn't help you.



This being said, you aren't the first to notice this issue. In fact, Ada was designed that dereferencing a pointer object is always assumed when a dot is used. In those cases where you want the actual pointer, you have to use .all. The ambiguity (pointer vs. pointed to object) is still there, but resolved by moving the extra syntax to the weirder case.






share|improve this answer


























  • +5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

    – Raffzahn
    3 hours ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









19














In the embryonic form of C described in the 1974 C Reference Manual, there was no requirement that the left operand of . actually be a structure, nor that the left operand of -> actually be a pointer. The -> operator meant "interpret the value of the left operand as a pointer, add the offset associated with the indicated structure member name, and dereference the resulting pointer as an object of the appropriate type. The . operator effectively took the address of the left operand and then applied ->.



Thus, given:



struct q { int x, y; };
int a[2];


the expressions a[0].y and a[0]->y would be interpreted in a fashion equivalent to ((struct q*)&a[0])->y and ((struct q*)a[0])->y, respectively.



If the compiler had examined the type of the left operand to the . operator, it could have used that to select between the two behaviors for it. It was probably easier, however, to have two operators whose behaviors didn't depend upon the left operand's type.






share|improve this answer



















  • 2





    To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

    – Raffzahn
    5 hours ago
















19














In the embryonic form of C described in the 1974 C Reference Manual, there was no requirement that the left operand of . actually be a structure, nor that the left operand of -> actually be a pointer. The -> operator meant "interpret the value of the left operand as a pointer, add the offset associated with the indicated structure member name, and dereference the resulting pointer as an object of the appropriate type. The . operator effectively took the address of the left operand and then applied ->.



Thus, given:



struct q { int x, y; };
int a[2];


the expressions a[0].y and a[0]->y would be interpreted in a fashion equivalent to ((struct q*)&a[0])->y and ((struct q*)a[0])->y, respectively.



If the compiler had examined the type of the left operand to the . operator, it could have used that to select between the two behaviors for it. It was probably easier, however, to have two operators whose behaviors didn't depend upon the left operand's type.






share|improve this answer



















  • 2





    To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

    – Raffzahn
    5 hours ago














19












19








19







In the embryonic form of C described in the 1974 C Reference Manual, there was no requirement that the left operand of . actually be a structure, nor that the left operand of -> actually be a pointer. The -> operator meant "interpret the value of the left operand as a pointer, add the offset associated with the indicated structure member name, and dereference the resulting pointer as an object of the appropriate type. The . operator effectively took the address of the left operand and then applied ->.



Thus, given:



struct q { int x, y; };
int a[2];


the expressions a[0].y and a[0]->y would be interpreted in a fashion equivalent to ((struct q*)&a[0])->y and ((struct q*)a[0])->y, respectively.



If the compiler had examined the type of the left operand to the . operator, it could have used that to select between the two behaviors for it. It was probably easier, however, to have two operators whose behaviors didn't depend upon the left operand's type.






share|improve this answer













In the embryonic form of C described in the 1974 C Reference Manual, there was no requirement that the left operand of . actually be a structure, nor that the left operand of -> actually be a pointer. The -> operator meant "interpret the value of the left operand as a pointer, add the offset associated with the indicated structure member name, and dereference the resulting pointer as an object of the appropriate type. The . operator effectively took the address of the left operand and then applied ->.



Thus, given:



struct q { int x, y; };
int a[2];


the expressions a[0].y and a[0]->y would be interpreted in a fashion equivalent to ((struct q*)&a[0])->y and ((struct q*)a[0])->y, respectively.



If the compiler had examined the type of the left operand to the . operator, it could have used that to select between the two behaviors for it. It was probably easier, however, to have two operators whose behaviors didn't depend upon the left operand's type.







share|improve this answer












share|improve this answer



share|improve this answer










answered 5 hours ago









supercatsupercat

8,405943




8,405943








  • 2





    To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

    – Raffzahn
    5 hours ago














  • 2





    To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

    – Raffzahn
    5 hours ago








2




2





To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

– Raffzahn
5 hours ago





To the point. As well as the last part about being 'easier'. C wasn'T designed to be a language as comfortable as possible, but to be translated as linear as possible. Resolving contextual information adds complexity and ambiguity. Nothing one wants t have when the task is to write an OS as close to the machine as possible while having the luxury of structured programming support.

– Raffzahn
5 hours ago











12














I think there are two factors that led to standardization of the distinct operator "->" for accessing data members using a pointer.




  1. You assume that the C compiler would recognize the type of the LHS as being a pointer. But programmers could, and often did, override the initial typing (variable declaration) by using a typecast.

  2. In order to make the code more readable and less prone to unintended side-effects, it is useful to distinguish operations using pointers.


A very common feature of idiomatic C code is that a structure passed to a function as a pointer is modified within the function. Thus, the result is returned implicitly, by the side-effect of the structure variable in the calling function having been modified by the callee. This sort of approach violates modern sensibilities about loosely coupled code, but it was a simple and efficient means of dealing with complexity in C code. I would say the programmer was greatly assisted in maintaining the readability of such code by having distinct operations that made it clear whether some (possibly shared) memory pointer was the thing whose target was being modified.






share|improve this answer


























  • Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

    – Dr Sheldon
    7 hours ago








  • 1





    @DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

    – Brian H
    6 hours ago


















12














I think there are two factors that led to standardization of the distinct operator "->" for accessing data members using a pointer.




  1. You assume that the C compiler would recognize the type of the LHS as being a pointer. But programmers could, and often did, override the initial typing (variable declaration) by using a typecast.

  2. In order to make the code more readable and less prone to unintended side-effects, it is useful to distinguish operations using pointers.


A very common feature of idiomatic C code is that a structure passed to a function as a pointer is modified within the function. Thus, the result is returned implicitly, by the side-effect of the structure variable in the calling function having been modified by the callee. This sort of approach violates modern sensibilities about loosely coupled code, but it was a simple and efficient means of dealing with complexity in C code. I would say the programmer was greatly assisted in maintaining the readability of such code by having distinct operations that made it clear whether some (possibly shared) memory pointer was the thing whose target was being modified.






share|improve this answer


























  • Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

    – Dr Sheldon
    7 hours ago








  • 1





    @DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

    – Brian H
    6 hours ago
















12












12








12







I think there are two factors that led to standardization of the distinct operator "->" for accessing data members using a pointer.




  1. You assume that the C compiler would recognize the type of the LHS as being a pointer. But programmers could, and often did, override the initial typing (variable declaration) by using a typecast.

  2. In order to make the code more readable and less prone to unintended side-effects, it is useful to distinguish operations using pointers.


A very common feature of idiomatic C code is that a structure passed to a function as a pointer is modified within the function. Thus, the result is returned implicitly, by the side-effect of the structure variable in the calling function having been modified by the callee. This sort of approach violates modern sensibilities about loosely coupled code, but it was a simple and efficient means of dealing with complexity in C code. I would say the programmer was greatly assisted in maintaining the readability of such code by having distinct operations that made it clear whether some (possibly shared) memory pointer was the thing whose target was being modified.






share|improve this answer















I think there are two factors that led to standardization of the distinct operator "->" for accessing data members using a pointer.




  1. You assume that the C compiler would recognize the type of the LHS as being a pointer. But programmers could, and often did, override the initial typing (variable declaration) by using a typecast.

  2. In order to make the code more readable and less prone to unintended side-effects, it is useful to distinguish operations using pointers.


A very common feature of idiomatic C code is that a structure passed to a function as a pointer is modified within the function. Thus, the result is returned implicitly, by the side-effect of the structure variable in the calling function having been modified by the callee. This sort of approach violates modern sensibilities about loosely coupled code, but it was a simple and efficient means of dealing with complexity in C code. I would say the programmer was greatly assisted in maintaining the readability of such code by having distinct operations that made it clear whether some (possibly shared) memory pointer was the thing whose target was being modified.







share|improve this answer














share|improve this answer



share|improve this answer








edited 6 hours ago

























answered 7 hours ago









Brian HBrian H

18.3k69158




18.3k69158













  • Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

    – Dr Sheldon
    7 hours ago








  • 1





    @DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

    – Brian H
    6 hours ago





















  • Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

    – Dr Sheldon
    7 hours ago








  • 1





    @DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

    – Brian H
    6 hours ago



















Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

– Dr Sheldon
7 hours ago







Could you give an example of your point #1? The type of the result of a typecast is well-defined... it's the type you are typecasting to. Should that be a pointer to a structure, the compiler has enough information to access its members. Whatever the type was prior to the typecast is irrelevant.

– Dr Sheldon
7 hours ago






1




1





@DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

– Brian H
6 hours ago







@DrSheldon Yes, you are correct. The compiler can check operators used against the type of a typecast too, and interpret a single operator appropriately. But then you are removing one of the compiler checks against programming errors. I think #1 and #2 actually work in concert to push the programmer toward care with pointer references. If the compiler tries to be "too smart" it ends up misinterpreting what the error-prone programmer intended, and (perhaps more important) makes the code harder to read.

– Brian H
6 hours ago













9














Despite your assertion, there would in fact be situations where it would be ambigious.



First off, early C compilers were very simple. This was in fact the main appeal of the language, as compilers for it were very easy to create and could run on very small systems, like early 16/32 microprocessors.



Adding a bunch of code for hitting all the niche cases of type inference would have drastically added to the amount of code required to make a C compiler. In fact, I've argued as a (half) joke that K&R C had type inference, but it always inferred int. If you didn't tell C what type an object was, it assumed int (which could cause some really gnarly bugs, let me tell you...)



Secondly, since K&R C was weakly(barely) typed, the information in many cases flat out wasn't available. The destination type of a pointer assignment can be an int, or visa versa, and K&R C has no problem with that. The compiler simply cannot infer a dereference. The coder is assumed to know what she's doing.



Also realize that in C pointers and arrays are essentially syntactic sugar for each other. This means now your . operator would have to automagically work on arrays too. For instance, if member happened to be a char array, now structure.member would return with the first character. And again, both chars and pointer are assignable into ints, so context doesn't help you.



This being said, you aren't the first to notice this issue. In fact, Ada was designed that dereferencing a pointer object is always assumed when a dot is used. In those cases where you want the actual pointer, you have to use .all. The ambiguity (pointer vs. pointed to object) is still there, but resolved by moving the extra syntax to the weirder case.






share|improve this answer


























  • +5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

    – Raffzahn
    3 hours ago
















9














Despite your assertion, there would in fact be situations where it would be ambigious.



First off, early C compilers were very simple. This was in fact the main appeal of the language, as compilers for it were very easy to create and could run on very small systems, like early 16/32 microprocessors.



Adding a bunch of code for hitting all the niche cases of type inference would have drastically added to the amount of code required to make a C compiler. In fact, I've argued as a (half) joke that K&R C had type inference, but it always inferred int. If you didn't tell C what type an object was, it assumed int (which could cause some really gnarly bugs, let me tell you...)



Secondly, since K&R C was weakly(barely) typed, the information in many cases flat out wasn't available. The destination type of a pointer assignment can be an int, or visa versa, and K&R C has no problem with that. The compiler simply cannot infer a dereference. The coder is assumed to know what she's doing.



Also realize that in C pointers and arrays are essentially syntactic sugar for each other. This means now your . operator would have to automagically work on arrays too. For instance, if member happened to be a char array, now structure.member would return with the first character. And again, both chars and pointer are assignable into ints, so context doesn't help you.



This being said, you aren't the first to notice this issue. In fact, Ada was designed that dereferencing a pointer object is always assumed when a dot is used. In those cases where you want the actual pointer, you have to use .all. The ambiguity (pointer vs. pointed to object) is still there, but resolved by moving the extra syntax to the weirder case.






share|improve this answer


























  • +5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

    – Raffzahn
    3 hours ago














9












9








9







Despite your assertion, there would in fact be situations where it would be ambigious.



First off, early C compilers were very simple. This was in fact the main appeal of the language, as compilers for it were very easy to create and could run on very small systems, like early 16/32 microprocessors.



Adding a bunch of code for hitting all the niche cases of type inference would have drastically added to the amount of code required to make a C compiler. In fact, I've argued as a (half) joke that K&R C had type inference, but it always inferred int. If you didn't tell C what type an object was, it assumed int (which could cause some really gnarly bugs, let me tell you...)



Secondly, since K&R C was weakly(barely) typed, the information in many cases flat out wasn't available. The destination type of a pointer assignment can be an int, or visa versa, and K&R C has no problem with that. The compiler simply cannot infer a dereference. The coder is assumed to know what she's doing.



Also realize that in C pointers and arrays are essentially syntactic sugar for each other. This means now your . operator would have to automagically work on arrays too. For instance, if member happened to be a char array, now structure.member would return with the first character. And again, both chars and pointer are assignable into ints, so context doesn't help you.



This being said, you aren't the first to notice this issue. In fact, Ada was designed that dereferencing a pointer object is always assumed when a dot is used. In those cases where you want the actual pointer, you have to use .all. The ambiguity (pointer vs. pointed to object) is still there, but resolved by moving the extra syntax to the weirder case.






share|improve this answer















Despite your assertion, there would in fact be situations where it would be ambigious.



First off, early C compilers were very simple. This was in fact the main appeal of the language, as compilers for it were very easy to create and could run on very small systems, like early 16/32 microprocessors.



Adding a bunch of code for hitting all the niche cases of type inference would have drastically added to the amount of code required to make a C compiler. In fact, I've argued as a (half) joke that K&R C had type inference, but it always inferred int. If you didn't tell C what type an object was, it assumed int (which could cause some really gnarly bugs, let me tell you...)



Secondly, since K&R C was weakly(barely) typed, the information in many cases flat out wasn't available. The destination type of a pointer assignment can be an int, or visa versa, and K&R C has no problem with that. The compiler simply cannot infer a dereference. The coder is assumed to know what she's doing.



Also realize that in C pointers and arrays are essentially syntactic sugar for each other. This means now your . operator would have to automagically work on arrays too. For instance, if member happened to be a char array, now structure.member would return with the first character. And again, both chars and pointer are assignable into ints, so context doesn't help you.



This being said, you aren't the first to notice this issue. In fact, Ada was designed that dereferencing a pointer object is always assumed when a dot is used. In those cases where you want the actual pointer, you have to use .all. The ambiguity (pointer vs. pointed to object) is still there, but resolved by moving the extra syntax to the weirder case.







share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago

























answered 3 hours ago









T.E.D.T.E.D.

70125




70125













  • +5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

    – Raffzahn
    3 hours ago



















  • +5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

    – Raffzahn
    3 hours ago

















+5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

– Raffzahn
3 hours ago





+5 for the 'syntactic sugar' and another +5 for showing a solution with the Ada reference.

– Raffzahn
3 hours ago


















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