Why does the fist print statement of b in the function print 5 and not 6?

Question

I'm confused. I would expect that the first print statement in the function would print 6 and not 5. Because a++ + c++ = (1+1) + (3+1) = (2+4 = 6).

#include <iostream>
#include <string>

int main()
{
    int a = 1;
    int b = 2;
    std::cout << "intitialisation b:"<< b << std::endl;
    int c = 3;
    {
        auto b = ++a + c++;
        std::cout << "increment b in function:" << b << std::endl;
        auto e = b;
        c += ++b;
        std::cout << "increment b in function:" << b << std::endl;
    }
    std::cout << "increment b out function:" << b << std::endl;
    int* p = &a;
    int* q = &b;
    std::cout <<"value pointer:" << *q << std::endl;
    ++(*q);
    std::cout <<"value pointer:" << *q << std::endl;
    *p += a++ + (*p)++;
}
 
intitialisation b:2
increment b in function:5
increment b in function:6
increment b out function:2
value pointer:2
value pointer:3

Nadine

Answer 1

The behaviour of the statement

*p += a++ + (*p)++;

is undefined. That's because you have simultaneous reads and writes to a (sometimes via the pointer) in an unsequenced step.

That means that the entire program is undefined, somewhat paradoxically perhaps, including any statements that, conceptually speaking, have already been ran prior to that one. And that includes the specific statement you're asking about!

Setting that aside, ++a evaluates to the incremented value of a, c++ evaluates to the unincremented value of c. So if there were no undefined constructs in your code, the output would be 2 + 3 which is 5.