++和&&的运算符优先级

Question

I have the following code

int a=0,b=5;
int c=a&&++b;
cout<<b;

When I run this snippet, the output value is 5 .

From my understanding, ++ has a greater operator precedence, and so shouldn't ++b be evaluated before the && leading to the value of b to be 6 ?

Why does the value in b not get incremented? Is it because I misunderstood the precedence or some other feature?

Answer 1

Precedence isn't what matters here.

&& does short circuit evaluation, so first its left operand is evaluated. Then if and only if that is non-zero, its right operand is evaluated.

In your case, its left operand is zero, so its right operand isn't evaluated. Thus, ++b is never evaluated, so the value of b isn't changed.

Answer 2

You misunderstand what operator precedence means in the language. It does not define order of evaluation but rather how expression syntaxically parsed.

Let's look into expression:

*ptr++ = b + c / d;

operator precedence means that it compiled as:

( *(ptr++) ) = ( b + ( c / d ) );

not:

( *( (ptr++) = b ) + c ) / d;

or something like this. But it does not define in which order they evaluated directly.

So back to original:

( *(ptr++) ) = ( b + ( c / d ) );

in this case we have operator= which requires output of operator* and output of operator+ so those operators have to be evaluated before assignment, operator+ requires output of operator/ so operator/ has to be evaluated before operator+ and so on. But because operator* has higher precedence than operator+ does not mean it will be evaluated first in this case. It's up to compiler to decide as there is no direct dependencies between them.