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

Question

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?

Yes i know but i am completely confused how it is working i also tried some different variants of the same code but still working. — Pankaj Prakash, Jun 17 '15 at 07:51
The unary + operator is usually rather confusing, as it has very limited practical use. It is mainly just there in the language to make it consistent with the far more useful unary - operator. — Lundin, Jun 17 '15 at 09:31

nneonneo · Answer 1 · 2015-06-17T07:54:34.847

5

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).

edited Jun 17 '15 at 07:54

answered Jun 17 '15 at 07:50

nneonneo

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and how this would evaluate sum = n - - - n + - n; – Pankaj Prakash Jun 17 '15 at 07:57
@PankajPrakash `--` will become a different operator without space because of maximal crunch rule. – Mohit Jain Jun 17 '15 at 07:59
@PankajPrakash Figure it out yourself - it's a good exercise. – nneonneo Jun 17 '15 at 07:59
@MohitJain: There are spaces, so it will parse as two unary operators. – nneonneo Jun 17 '15 at 08:00
@nneonneo OK, spaces are hard to spot – Mohit Jain Jun 17 '15 at 08:01
int sum = n - - - n + - n; is evaluating to -2. While when the same line with parenthesis int sum = n - (- (- (n + (- n)))); evaluates to 2. – Pankaj Prakash Jun 17 '15 at 08:02
Then you have the wrong evaluation. Note that unary operators bind "tighter" than binary operators. Try again! – nneonneo Jun 17 '15 at 08:03
@nneonneo wouldn't int sum = n - - - n + - n; is evaluated as int sum = n - (- (- (n + (- n)))); – Pankaj Prakash Jun 17 '15 at 08:05

score 4 · Accepted Answer · answered Jun 17 '15 at 07:51

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.

Yu Hao · Answer 3 · 2015-06-17T07:57:59.520

3

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.

edited Jun 17 '15 at 07:57

answered Jun 17 '15 at 07:50

Yu Hao

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

3 Answers3