bahrep - 5 months ago 14

C Question

I'm learning C basics and I'm trying to understand why the output of this example is

`1 2 1 1`

`int main()`

{

int a, b, c, d;

a = b = c = d = 1;

a = ++b > 1 || ++c > 1 && ++d > 1;

printf("%d%d%d%d", a, b, c, d);

}

- As far as I understand, only the left operand before Logical OR is getting evaluated because it is True and there is no need to evaluate the right one. The right operand is not evaluated. Is this right?
`||`

- The left operand is true because equals
`++b > 1`

(2 is larger than 1) and`1`

. Right?`a = 1`

Answer

The left side of the `||`

evaluates to `true`

, and therefore the right side does not need to be evaluated. Hence, `c`

and `d`

remain `1`

, `b`

is incremented to `2`

, and `a`

gets assigned to `true`

, which is promoted to `1`

.

**Explanation:**

```
a = b = c = d = 1;
```

All four variables equal `1`

.

```
a = ++b > 1
```

The variable `b`

gets incremented to `2`

*before* the rest of the RHS gets evaluated. This results in the expression `2 > 1`

, which is true. As others have mentioned, your `||`

expression is then *short-circuited*, which means that what follows the `||`

will not be evaluated because the program already knows that the entire RHS will be true regardless of what happens. This is why I did not bother to even write the expression `++c > 1 && ++d > 1`

, because it will be skipped.

And the variable `a`

is set to true, which is the same as `1`

.

Reference on promoting true to `int`

: Can I assume (bool)true == (int)1 for any C++ compiler?