What is the difference between XOR in parenthesis and without in C language

Question

i have some code and it checking chechsum by XOR on 5 first numbers, and it should be equal to the sixth number. But, what is the difference between

if (n0^n1^n2^n3^n4==n5)  return true;
    else return false;

and

if ((n0^n1^n2^n3^n4)==n5)  return true;
    else return false;

?

Because the first one is working and the second one is not.

Answer 1

The C grammar establishes operator precedence with == having higher precedence than ^ . So n0^n1^n2^n3^n4==n5 is equivalent to n0^n1^n2^n3^(n4==n5) , which differs from (n0^n1^n2^n3^n4)==n5 .

This is regarded by Kernighan and Ritchie and others as a mistake in the design of the C language, which occurred due to the history of its development.

Answer 2

C Operator Precedence of operator == is higher than the precedence of the operator ^ . Therefore in the first case the C compiler first executes n4==n5 which either results in 1 or 0 and then it will do all xor operators between n0..n3 and the result of the comparison.

In the second case the precedence of the () is higher than '=='. So, all xor operators will be done first and the comparison will be done the last.

It explains the difference in your program results.