This afternoon, I really don't know what I was doing with Operators and C. Eventually, I wrote some code which I was thinking wouldn't compile, But I don't know how it worked.
The code is:
#include <stdio.h>
int main()
{
int n=2;
int sum = n + - + - + - + n; /* This line */
printf("%d\n", sum);
return 0;
}
And the output is:
0
I am completely confused how the code compiled and what is happening behind the scene.
How does the line int sum = n + - + - + - + n; work?
All but the first are just unary operators.
n + - + - + - + n
is equivalent to
n + (-(+(-(+(-(+n))))))
which is in turn simply equal to
n + (-n)
after resolving all the unary operators.
-n is, of course, ordinary negation; +n does essentially nothing (though it has the side effect of forcing integral promotion).
int sum = n + - + - + - + n;
/* b u u u u u u */
/* Order: 7 6 5 4 3 2 1 */
is equivalent to:
n + (-(+(-(+(-(+n))))));
or simply n + (-n)
Note that unary operators bind more tightly than binary operators in C opeartor precedance table and associativity of unary operator +- is from right to left while of binary +- operators in from left to right.
Both of + and - are also unary operators. The result of +n is the (promoted) value of n. The result of -n is the negative of (promoted) n.
n + - + - + - + n;
is equivalent to:
n + (-(+(-(+(-(+n))))))
which is basically n + (-n) assuming no overflow happens.