Why is x == (x = y) not the same as (x = y) == x?











up vote
15
down vote

favorite
2












Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls







share|improve this question




















  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago















up vote
15
down vote

favorite
2












Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls







share|improve this question




















  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago













up vote
15
down vote

favorite
2









up vote
15
down vote

favorite
2






2





Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls







share|improve this question















Consider the following example:



class Quirky {

    public static void main(String args) {

        int x = 1;

        int y = 3;



        System.out.println(x == (x = y)); // false

        x = 1; // reset

        System.out.println((x = y) == x); // true

     }

}


I'm not sure if there is an item in the Java Language Specification that dictates loading the previous value of a variable for comparison with the right side (x = y) which, by the order implied by brackets, should be calculated first.



Why does the first expression evaluate to false, but the second evaluate to true? I would have expected (x = y) to be evaluated first, and then it would compare x with itself (3) and return true.






java variable-assignment operator-precedence jls




java variable-assignment operator-precedence jls






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago









Boann

36.6k1287121




36.6k1287121










asked 10 hours ago









John McClane

452114




452114








  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago














  • 14




    The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
    – Louis Wasserman
    10 hours ago












  • @LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
    – John McClane
    10 hours ago








  • 3




    Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
    – Louis Wasserman
    10 hours ago






  • 8




    Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
    – jpmc26
    7 hours ago








  • 1




    The real question is why you want to write code like this.
    – Broman
    2 hours ago








14




14




The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
10 hours ago






The left hand side is always evaluated before the right hand side. The brackets don't make a difference to that.
– Louis Wasserman
10 hours ago














@LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
– John McClane
10 hours ago






@LouisWasserman The term evaluate is, in my opinion, inapplicable here because x does not need to be evaluated, it just loaded from memory.
– John McClane
10 hours ago






3




3




Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
– Louis Wasserman
10 hours ago




Evaluating the expression x = y is certainly relevant, and causes the side effect that x is set to the value of y.
– Louis Wasserman
10 hours ago




8




8




Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
7 hours ago






Do yourself and your teammates a favor and don't mix state mutation into the same line as state examination. Doing so drastically reduces the readability of your code. (There are some cases where it's absolutely necessary because of atomicity requirements, but functions for those already exist and their purpose would be instantly recognized.)
– jpmc26
7 hours ago






1




1




The real question is why you want to write code like this.
– Broman
2 hours ago




The real question is why you want to write code like this.
– Broman
2 hours ago












3 Answers
3






active

oldest

votes

















up vote
26
down vote













== is a binary equality operator.




The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







share|improve this answer



















  • 1




    @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
    – Andrew Tobilko
    8 hours ago




















up vote
15
down vote













As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



//the example values

x = 1;

y = 2;


So in quickReplaceAndCompare the following is done:



x == (x = y)

1 == (x = y)

1 == (x = 2) //assign 2 to x, returns 2

1 == 2

false


and in badReplaceAndCompare():



(x = y) == x

(x = 2) == x //assign 2 to x, returns 2

2 == x

2 == 2

true


note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






share|improve this answer




























    up vote
    0
    down vote













    In the first test you're checking does 1 == 3.



    In the second test your checking does 3 == 3.



    (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



    In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






    share|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      26
      down vote













      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







      share|improve this answer



















      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago

















      up vote
      26
      down vote













      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







      share|improve this answer



















      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago















      up vote
      26
      down vote










      up vote
      26
      down vote









      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First







      share|improve this answer














      == is a binary equality operator.




      The left-hand operand of a binary operator appears to be fully evaluated before any part of the right-hand operand is evaluated.



      Java 11 Specification > Evaluation Order > Evaluate Left-Hand Operand First








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 9 hours ago

























      answered 9 hours ago









      Andrew Tobilko

      24.4k84182




      24.4k84182








      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago
















      • 1




        @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
        – Andrew Tobilko
        8 hours ago










      1




      1




      @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
      – Andrew Tobilko
      8 hours ago






      @JohnMcClane "by the order implied by brackets" - the parentheses don't set any order, they form an operand for ==. Without the parentheses, neither of these expressions would be syntactically correct.
      – Andrew Tobilko
      8 hours ago














      up vote
      15
      down vote













      As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



      //the example values
      
x = 1;
      
y = 2;


      So in quickReplaceAndCompare the following is done:



      x == (x = y)
      
1 == (x = y)
      
1 == (x = 2) //assign 2 to x, returns 2
      
1 == 2
      
false


      and in badReplaceAndCompare():



      (x = y) == x
      
(x = 2) == x //assign 2 to x, returns 2
      
2 == x
      
2 == 2
      
true


      note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






      share|improve this answer

























        up vote
        15
        down vote













        As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



        //the example values
        
x = 1;
        
y = 2;


        So in quickReplaceAndCompare the following is done:



        x == (x = y)
        
1 == (x = y)
        
1 == (x = 2) //assign 2 to x, returns 2
        
1 == 2
        
false


        and in badReplaceAndCompare():



        (x = y) == x
        
(x = 2) == x //assign 2 to x, returns 2
        
2 == x
        
2 == 2
        
true


        note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






        share|improve this answer























          up vote
          15
          down vote










          up vote
          15
          down vote









          As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



          //the example values
          
x = 1;
          
y = 2;


          So in quickReplaceAndCompare the following is done:



          x == (x = y)
          
1 == (x = y)
          
1 == (x = 2) //assign 2 to x, returns 2
          
1 == 2
          
false


          and in badReplaceAndCompare():



          (x = y) == x
          
(x = 2) == x //assign 2 to x, returns 2
          
2 == x
          
2 == 2
          
true


          note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.






          share|improve this answer












          As LouisWasserman sad, the expression is evaluated left to right. And java doesn't care what "evaluate" actually does, it only cares about generating a (non volatile, final) value to work with.



          //the example values
          
x = 1;
          
y = 2;


          So in quickReplaceAndCompare the following is done:



          x == (x = y)
          
1 == (x = y)
          
1 == (x = 2) //assign 2 to x, returns 2
          
1 == 2
          
false


          and in badReplaceAndCompare():



          (x = y) == x
          
(x = 2) == x //assign 2 to x, returns 2
          
2 == x
          
2 == 2
          
true


          note that badReplaceAndCompare() will always return true, because you are effectively comparing the assignment of a value to the variable it is assigned to, and a = b and b will, evaluated in that order, always be the same by definition.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 9 hours ago









          Poohl

          33319




          33319






















              up vote
              0
              down vote













              In the first test you're checking does 1 == 3.



              In the second test your checking does 3 == 3.



              (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



              In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






              share|improve this answer

























                up vote
                0
                down vote













                In the first test you're checking does 1 == 3.



                In the second test your checking does 3 == 3.



                (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



                In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  In the first test you're checking does 1 == 3.



                  In the second test your checking does 3 == 3.



                  (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



                  In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?






                  share|improve this answer












                  In the first test you're checking does 1 == 3.



                  In the second test your checking does 3 == 3.



                  (x = y) assigns the value and that value is tested. In the former example x = 1 first then x is assigned 3. Does 1 == 3?



                  In the latter, x is assigned 3, and obviously it's still 3. Does 3 == 3?







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  Michael Puckett II

                  3,67541436




                  3,67541436






























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                      南乌拉尔铁路局

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                      南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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