How does `int sum = n + - + - + - + n` compile where `n` is an `int`?

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?

--------------Solutions-------------

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.

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).

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.

Category:c# Time:2018-12-25 Views:1

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