Switch and do-while in c++

Question

#include <iostream>

int main() {
    int n = 3, i=0;

    switch (n % 2) 
    {
    case 0: 
        do {
            ++i;
    case 1:
            ++i;
        } while (--n > 0);
    }
    std::cout << i;
}

Why is the output 5 and not 6 ? Can someone explain it to me more detailed ?

Answer 1

Basically, when you enter the 'switch' you jump inside the loop body. Then the loop is executed ignoring the label 'case 1'.

It is like executing the following lines of code.

++i;
--n;
++i;
++i;
--n;
++i;
++i;
--n;

I never suggest a code style like that, because it is harder to maintain. I prefer instead a structured code. The style of your example reminds me the C64 Basic or Assembly.

While this code style is deprecable, the C language (C++ too) is flexible enough to allow you this. Take care because jumping randomly around could decrease the optimizer performance. At the end your code could run slower than the structured solution.

Answer 2

as n%2 = 1 then after the switch statement yo go to the case 1 which is in do while, so you entered in the do while then these lines will be exectued

++i // then i = 1
--n // now n = 2 which is >0
++i // i = 2
++i // i = 3
--n // now n = 1 which is > 0
++i // i = 4
++i // i = 5
--n //  now n = 0 and does not satisfy the condition so the loop breaks and i = 5