# 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?

```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# Views:11 Time:2018-12-15