Convert decimal to binary
This program converts a decimal number to a binary number. This is one of my first C programs and I am wondering if I have used the elements of this language properly. Suggestions for improvement are welcome.
#include <stdio.h>
#include <string.h>
void print_out_reversed(char string)
{
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
putchar('n');
}
void print_decimal_number_binary(int number)
{
if (number == 0)
{
printf("0n");
return;
}
char bits[sizeof(int) * 8 + 1] = {0};
int index = 0;
while (number > 0)
{
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
number = number / 2;
index++;
}
print_out_reversed(bits);
}
int main()
{
printf("enter number: ");
int number;
scanf("%i", &number);
print_decimal_number_binary(number);
}
beginner c
New contributor
add a comment |
This program converts a decimal number to a binary number. This is one of my first C programs and I am wondering if I have used the elements of this language properly. Suggestions for improvement are welcome.
#include <stdio.h>
#include <string.h>
void print_out_reversed(char string)
{
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
putchar('n');
}
void print_decimal_number_binary(int number)
{
if (number == 0)
{
printf("0n");
return;
}
char bits[sizeof(int) * 8 + 1] = {0};
int index = 0;
while (number > 0)
{
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
number = number / 2;
index++;
}
print_out_reversed(bits);
}
int main()
{
printf("enter number: ");
int number;
scanf("%i", &number);
print_decimal_number_binary(number);
}
beginner c
New contributor
add a comment |
This program converts a decimal number to a binary number. This is one of my first C programs and I am wondering if I have used the elements of this language properly. Suggestions for improvement are welcome.
#include <stdio.h>
#include <string.h>
void print_out_reversed(char string)
{
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
putchar('n');
}
void print_decimal_number_binary(int number)
{
if (number == 0)
{
printf("0n");
return;
}
char bits[sizeof(int) * 8 + 1] = {0};
int index = 0;
while (number > 0)
{
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
number = number / 2;
index++;
}
print_out_reversed(bits);
}
int main()
{
printf("enter number: ");
int number;
scanf("%i", &number);
print_decimal_number_binary(number);
}
beginner c
New contributor
This program converts a decimal number to a binary number. This is one of my first C programs and I am wondering if I have used the elements of this language properly. Suggestions for improvement are welcome.
#include <stdio.h>
#include <string.h>
void print_out_reversed(char string)
{
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
putchar('n');
}
void print_decimal_number_binary(int number)
{
if (number == 0)
{
printf("0n");
return;
}
char bits[sizeof(int) * 8 + 1] = {0};
int index = 0;
while (number > 0)
{
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
number = number / 2;
index++;
}
print_out_reversed(bits);
}
int main()
{
printf("enter number: ");
int number;
scanf("%i", &number);
print_decimal_number_binary(number);
}
beginner c
beginner c
New contributor
New contributor
edited 4 hours ago
New contributor
asked 4 hours ago
Vengeancos
564
564
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Terminology
It's important to be able to understand (and describe) what's actually going on. Your program
- converts from an integer decimal string representation to an integer using
scanf
. This integer is then represented as a binary number in the processor. - converts from that integer back into a string representation, but rather than it being decimal, it's binary.
So yes - it technically converts from "decimal to binary", but really it's "decimal string to integer to binary string".
Use const
void print_out_reversed(char string)
doesn't modify string
, so write const char string
.
Simplify your strlen
usage
This:
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
can be
for (int i = strlen(string)-1; i >= 0; i--)
putchar(string[i]);
It seems that you don't trust what strlen
is doing, which is why you have that intermediate while
loop. But that loop won't have any effect, because the null terminator will always be where strlen
says it is.
Use math instead of if
This:
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
can be
bits[index] = '0' + (number & 1);
Use combined operation and assignment
This:
number = number / 2;
should be
number /= 2;
or, for speed (which the compiler will do for you anyway)
number >>= 1;
add a comment |
#include stdio.h
#include string.h
void get_bits(void * num, char * out, int bytes);
int main(void)
{
int x = 0;
printf("Enter an integer: ");
scanf("%i", &x);
char bits[65] = {0};
get_bits(&x, bits, 4);
printf("%d in binary %sn", x, bits);
return 0;
}
//assumes char array of length 1 greater than
//number of bits
void get_bits(void * num, char *out, int bytes)
{
unsigned long long filter = 0x8000000000000000;
unsigned long long temp;
if(bytes <= 0) return;
if(bytes > 8) bytes = 8;
filter = filter >> (8*(sizeof(unsigned long long)-bytes));
memcpy(&temp, num, bytes);
int bits = 8*bytes;
for(int i=0;i<bits;i++) {
if(filter & temp)
out[i] = '1';
else
out[i] = '0';
temp = temp << 1;
}
out[bits] = '';
}
The code above stores characters representing the bits for char, short, int, or long long in a character array (up to 64 bits). It does not require reversing the order of bits and restoring them. Essentially, it creates a mask ("filter") with a single bit set. This is always the highest order bit. initially, it is the highest order bit for a 64 bit number. For a 32-bit integer, the mask will be shifted to have the 32 bit set, all others 0. The mask will be adjusted for whatever number of bytes the caller indicates are in the number.
A bit-wise and is performed with the number. If the corresponding bit is 1 in the number, it stores a 1 character in the array passed to get_bits. The number, temporarily stored in the variable, temp, is shifted left in each iteration of the for loop. This allows comparison each iteration of the highest order bit in the number. If it is not set in the number, filter & temp is 0 and a 0 character is stored at that position in the output string.
The user of the function get_bits must pass a void pointer, which stores the address of the number, for which the bits are determined. The argument out is a pointer to a character array for the output string. This should be one character longer than the number of bits in the number. The argument bytes represents the bytes of data required to store the number. The output character array is null terminated at position corresponding to the number of bits in the number (that is the length of the string with null terminator is one greater than the number of bytes *8).
New contributor
Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Terminology
It's important to be able to understand (and describe) what's actually going on. Your program
- converts from an integer decimal string representation to an integer using
scanf
. This integer is then represented as a binary number in the processor. - converts from that integer back into a string representation, but rather than it being decimal, it's binary.
So yes - it technically converts from "decimal to binary", but really it's "decimal string to integer to binary string".
Use const
void print_out_reversed(char string)
doesn't modify string
, so write const char string
.
Simplify your strlen
usage
This:
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
can be
for (int i = strlen(string)-1; i >= 0; i--)
putchar(string[i]);
It seems that you don't trust what strlen
is doing, which is why you have that intermediate while
loop. But that loop won't have any effect, because the null terminator will always be where strlen
says it is.
Use math instead of if
This:
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
can be
bits[index] = '0' + (number & 1);
Use combined operation and assignment
This:
number = number / 2;
should be
number /= 2;
or, for speed (which the compiler will do for you anyway)
number >>= 1;
add a comment |
Terminology
It's important to be able to understand (and describe) what's actually going on. Your program
- converts from an integer decimal string representation to an integer using
scanf
. This integer is then represented as a binary number in the processor. - converts from that integer back into a string representation, but rather than it being decimal, it's binary.
So yes - it technically converts from "decimal to binary", but really it's "decimal string to integer to binary string".
Use const
void print_out_reversed(char string)
doesn't modify string
, so write const char string
.
Simplify your strlen
usage
This:
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
can be
for (int i = strlen(string)-1; i >= 0; i--)
putchar(string[i]);
It seems that you don't trust what strlen
is doing, which is why you have that intermediate while
loop. But that loop won't have any effect, because the null terminator will always be where strlen
says it is.
Use math instead of if
This:
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
can be
bits[index] = '0' + (number & 1);
Use combined operation and assignment
This:
number = number / 2;
should be
number /= 2;
or, for speed (which the compiler will do for you anyway)
number >>= 1;
add a comment |
Terminology
It's important to be able to understand (and describe) what's actually going on. Your program
- converts from an integer decimal string representation to an integer using
scanf
. This integer is then represented as a binary number in the processor. - converts from that integer back into a string representation, but rather than it being decimal, it's binary.
So yes - it technically converts from "decimal to binary", but really it's "decimal string to integer to binary string".
Use const
void print_out_reversed(char string)
doesn't modify string
, so write const char string
.
Simplify your strlen
usage
This:
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
can be
for (int i = strlen(string)-1; i >= 0; i--)
putchar(string[i]);
It seems that you don't trust what strlen
is doing, which is why you have that intermediate while
loop. But that loop won't have any effect, because the null terminator will always be where strlen
says it is.
Use math instead of if
This:
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
can be
bits[index] = '0' + (number & 1);
Use combined operation and assignment
This:
number = number / 2;
should be
number /= 2;
or, for speed (which the compiler will do for you anyway)
number >>= 1;
Terminology
It's important to be able to understand (and describe) what's actually going on. Your program
- converts from an integer decimal string representation to an integer using
scanf
. This integer is then represented as a binary number in the processor. - converts from that integer back into a string representation, but rather than it being decimal, it's binary.
So yes - it technically converts from "decimal to binary", but really it's "decimal string to integer to binary string".
Use const
void print_out_reversed(char string)
doesn't modify string
, so write const char string
.
Simplify your strlen
usage
This:
int index = strlen(string);
while (string[index] != '')
index--;
for (int i = index; i >= 0; i--)
putchar(string[i]);
can be
for (int i = strlen(string)-1; i >= 0; i--)
putchar(string[i]);
It seems that you don't trust what strlen
is doing, which is why you have that intermediate while
loop. But that loop won't have any effect, because the null terminator will always be where strlen
says it is.
Use math instead of if
This:
if (number % 2 == 0)
{
bits[index] = '0';
}
else
{
bits[index] = '1';
}
can be
bits[index] = '0' + (number & 1);
Use combined operation and assignment
This:
number = number / 2;
should be
number /= 2;
or, for speed (which the compiler will do for you anyway)
number >>= 1;
answered 4 hours ago
Reinderien
3,733721
3,733721
add a comment |
add a comment |
#include stdio.h
#include string.h
void get_bits(void * num, char * out, int bytes);
int main(void)
{
int x = 0;
printf("Enter an integer: ");
scanf("%i", &x);
char bits[65] = {0};
get_bits(&x, bits, 4);
printf("%d in binary %sn", x, bits);
return 0;
}
//assumes char array of length 1 greater than
//number of bits
void get_bits(void * num, char *out, int bytes)
{
unsigned long long filter = 0x8000000000000000;
unsigned long long temp;
if(bytes <= 0) return;
if(bytes > 8) bytes = 8;
filter = filter >> (8*(sizeof(unsigned long long)-bytes));
memcpy(&temp, num, bytes);
int bits = 8*bytes;
for(int i=0;i<bits;i++) {
if(filter & temp)
out[i] = '1';
else
out[i] = '0';
temp = temp << 1;
}
out[bits] = '';
}
The code above stores characters representing the bits for char, short, int, or long long in a character array (up to 64 bits). It does not require reversing the order of bits and restoring them. Essentially, it creates a mask ("filter") with a single bit set. This is always the highest order bit. initially, it is the highest order bit for a 64 bit number. For a 32-bit integer, the mask will be shifted to have the 32 bit set, all others 0. The mask will be adjusted for whatever number of bytes the caller indicates are in the number.
A bit-wise and is performed with the number. If the corresponding bit is 1 in the number, it stores a 1 character in the array passed to get_bits. The number, temporarily stored in the variable, temp, is shifted left in each iteration of the for loop. This allows comparison each iteration of the highest order bit in the number. If it is not set in the number, filter & temp is 0 and a 0 character is stored at that position in the output string.
The user of the function get_bits must pass a void pointer, which stores the address of the number, for which the bits are determined. The argument out is a pointer to a character array for the output string. This should be one character longer than the number of bits in the number. The argument bytes represents the bytes of data required to store the number. The output character array is null terminated at position corresponding to the number of bits in the number (that is the length of the string with null terminator is one greater than the number of bytes *8).
New contributor
Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
add a comment |
#include stdio.h
#include string.h
void get_bits(void * num, char * out, int bytes);
int main(void)
{
int x = 0;
printf("Enter an integer: ");
scanf("%i", &x);
char bits[65] = {0};
get_bits(&x, bits, 4);
printf("%d in binary %sn", x, bits);
return 0;
}
//assumes char array of length 1 greater than
//number of bits
void get_bits(void * num, char *out, int bytes)
{
unsigned long long filter = 0x8000000000000000;
unsigned long long temp;
if(bytes <= 0) return;
if(bytes > 8) bytes = 8;
filter = filter >> (8*(sizeof(unsigned long long)-bytes));
memcpy(&temp, num, bytes);
int bits = 8*bytes;
for(int i=0;i<bits;i++) {
if(filter & temp)
out[i] = '1';
else
out[i] = '0';
temp = temp << 1;
}
out[bits] = '';
}
The code above stores characters representing the bits for char, short, int, or long long in a character array (up to 64 bits). It does not require reversing the order of bits and restoring them. Essentially, it creates a mask ("filter") with a single bit set. This is always the highest order bit. initially, it is the highest order bit for a 64 bit number. For a 32-bit integer, the mask will be shifted to have the 32 bit set, all others 0. The mask will be adjusted for whatever number of bytes the caller indicates are in the number.
A bit-wise and is performed with the number. If the corresponding bit is 1 in the number, it stores a 1 character in the array passed to get_bits. The number, temporarily stored in the variable, temp, is shifted left in each iteration of the for loop. This allows comparison each iteration of the highest order bit in the number. If it is not set in the number, filter & temp is 0 and a 0 character is stored at that position in the output string.
The user of the function get_bits must pass a void pointer, which stores the address of the number, for which the bits are determined. The argument out is a pointer to a character array for the output string. This should be one character longer than the number of bits in the number. The argument bytes represents the bytes of data required to store the number. The output character array is null terminated at position corresponding to the number of bits in the number (that is the length of the string with null terminator is one greater than the number of bytes *8).
New contributor
Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
add a comment |
#include stdio.h
#include string.h
void get_bits(void * num, char * out, int bytes);
int main(void)
{
int x = 0;
printf("Enter an integer: ");
scanf("%i", &x);
char bits[65] = {0};
get_bits(&x, bits, 4);
printf("%d in binary %sn", x, bits);
return 0;
}
//assumes char array of length 1 greater than
//number of bits
void get_bits(void * num, char *out, int bytes)
{
unsigned long long filter = 0x8000000000000000;
unsigned long long temp;
if(bytes <= 0) return;
if(bytes > 8) bytes = 8;
filter = filter >> (8*(sizeof(unsigned long long)-bytes));
memcpy(&temp, num, bytes);
int bits = 8*bytes;
for(int i=0;i<bits;i++) {
if(filter & temp)
out[i] = '1';
else
out[i] = '0';
temp = temp << 1;
}
out[bits] = '';
}
The code above stores characters representing the bits for char, short, int, or long long in a character array (up to 64 bits). It does not require reversing the order of bits and restoring them. Essentially, it creates a mask ("filter") with a single bit set. This is always the highest order bit. initially, it is the highest order bit for a 64 bit number. For a 32-bit integer, the mask will be shifted to have the 32 bit set, all others 0. The mask will be adjusted for whatever number of bytes the caller indicates are in the number.
A bit-wise and is performed with the number. If the corresponding bit is 1 in the number, it stores a 1 character in the array passed to get_bits. The number, temporarily stored in the variable, temp, is shifted left in each iteration of the for loop. This allows comparison each iteration of the highest order bit in the number. If it is not set in the number, filter & temp is 0 and a 0 character is stored at that position in the output string.
The user of the function get_bits must pass a void pointer, which stores the address of the number, for which the bits are determined. The argument out is a pointer to a character array for the output string. This should be one character longer than the number of bits in the number. The argument bytes represents the bytes of data required to store the number. The output character array is null terminated at position corresponding to the number of bits in the number (that is the length of the string with null terminator is one greater than the number of bytes *8).
New contributor
#include stdio.h
#include string.h
void get_bits(void * num, char * out, int bytes);
int main(void)
{
int x = 0;
printf("Enter an integer: ");
scanf("%i", &x);
char bits[65] = {0};
get_bits(&x, bits, 4);
printf("%d in binary %sn", x, bits);
return 0;
}
//assumes char array of length 1 greater than
//number of bits
void get_bits(void * num, char *out, int bytes)
{
unsigned long long filter = 0x8000000000000000;
unsigned long long temp;
if(bytes <= 0) return;
if(bytes > 8) bytes = 8;
filter = filter >> (8*(sizeof(unsigned long long)-bytes));
memcpy(&temp, num, bytes);
int bits = 8*bytes;
for(int i=0;i<bits;i++) {
if(filter & temp)
out[i] = '1';
else
out[i] = '0';
temp = temp << 1;
}
out[bits] = '';
}
The code above stores characters representing the bits for char, short, int, or long long in a character array (up to 64 bits). It does not require reversing the order of bits and restoring them. Essentially, it creates a mask ("filter") with a single bit set. This is always the highest order bit. initially, it is the highest order bit for a 64 bit number. For a 32-bit integer, the mask will be shifted to have the 32 bit set, all others 0. The mask will be adjusted for whatever number of bytes the caller indicates are in the number.
A bit-wise and is performed with the number. If the corresponding bit is 1 in the number, it stores a 1 character in the array passed to get_bits. The number, temporarily stored in the variable, temp, is shifted left in each iteration of the for loop. This allows comparison each iteration of the highest order bit in the number. If it is not set in the number, filter & temp is 0 and a 0 character is stored at that position in the output string.
The user of the function get_bits must pass a void pointer, which stores the address of the number, for which the bits are determined. The argument out is a pointer to a character array for the output string. This should be one character longer than the number of bits in the number. The argument bytes represents the bytes of data required to store the number. The output character array is null terminated at position corresponding to the number of bits in the number (that is the length of the string with null terminator is one greater than the number of bytes *8).
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RJM
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Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
add a comment |
Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
Welcome to Code Review! You have presented an alternative solution, but haven't reviewed the code. Please explain your reasoning (how your solution works and why it is better than the original) so that the author and other readers can learn from your thought process. Please read Why are alternative solutions not welcome?
– Sᴀᴍ Onᴇᴌᴀ
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ Thanks. New on this site. I was having trouble just editing the code. I will put it in.
– RJM
2 hours ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
@SᴀᴍOnᴇᴌᴀ So, different than Math Stack Exchange, it seems we should not put up alternatives. Is that so? I thought my answer was so different, that it provides extended understanding of how to answer.
– RJM
1 hour ago
add a comment |
Vengeancos is a new contributor. Be nice, and check out our Code of Conduct.
Vengeancos is a new contributor. Be nice, and check out our Code of Conduct.
Vengeancos is a new contributor. Be nice, and check out our Code of Conduct.
Vengeancos is a new contributor. Be nice, and check out our Code of Conduct.
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