comma separated assignment to an integer variable n c [duplicate] - c

Question

This question already has answers here:
What does the comma operator , do?
(8 answers)
Closed 7 years ago.
I'm unable to understand the working of this piece of code.
#include<stdio.h>
void main(){
int a,b;
a=3,1;
b=(5,4);
printf("%d",a+b);
}
The output is 7. What is that assignment?

Answer 1

Comma operator evaluates its first operand and discards the result, and then evaluates the second operand and returns this value.
After executing
a=3,1; // (a = 3), 1;
a will have 3 and after
b=(5,4); // Discard 5 and the value of the expression (5,4) will be 4
b will have 4.
Some more examples on Wikipedia:
// Examples: Descriptions: Values after line is evaluated:
int a=1, b=2, c=3, i=0; // commas act as separators in this line, not as an operator
// ... a=1, b=2, c=3, i=0
i = (a, b); // stores b into i
// ... a=1, b=2, c=3, i=2
i = a, b; // stores a into i. Equivalent to (i = a), b;
// ... a=1, b=2, c=3, i=1
i = (a += 2, a + b); // increases a by 2, then stores a+b = 3+2 into i
// ... a=3, b=2, c=3, i=5
i = a += 2, a + b; // increases a by 2, then stores a to i, and discards unused
// a + b rvalue. Equivalent to (i = (a += 2)), a + b;
// ... a=5, b=2, c=3, i=5
i = a, b, c; // stores a into i, discarding the unused b and c rvalues
// ... a=5, b=2, c=3, i=5
i = (a, b, c); // stores c into i, discarding the unused a and b rvalues
// ... a=5, b=2, c=3, i=3
return a=4, b=5, c=6; // returns 6, not 4, since comma operator sequence points
// following the keyword 'return' are considered a single
// expression evaluating to rvalue of final subexpression c=6
return 1, 2, 3; // returns 3, not 1, for same reason as previous example
return(1), 2, 3; // returns 3, not 1, still for same reason as above. This
// example works as it does because return is a keyword, not
// a function call. Even though most compilers will allow for
// the construct return(value), the parentheses are syntactic
// sugar that get stripped out without syntactic analysis

Answer 2

a=3,1; // (a=3),1 -- value of expression is 1, side effect is changing a to 3
b=(5,4);
printf("%d",a+b); // 3 + 4

Answer 3

In this statement
a=3,1;
two operators are used: the assignment operator and the comma operator. The priority of the assignment operator is greater than the priority of the comma operator, thus this statement is equivalent to
( a = 3 ), 1;
1 is simply discarded, so a is assigned the value 3.
In this statement
b=(5,4);
due to the parentheses the comma operator is evaluated first. Its value is the value of the last expression, 4. So b is assigned the value 4.
As result you get a + b => 3 + 4 which equals 7.