JuniorTheoryCommon

What does the program print?

C++

#include <iostream>

int main() {
    int n = 2;
    switch (n) {
        case 1: std::cout << 'a'; break;
        case 2: std::cout << 'b';
        case 3: std::cout << 'c';
        default: std::cout << 'd';
    }
}

bcd

Expecting b: a case label does not end the code above it, so without break execution runs on into case 3 and further.
Expecting bc: falling through does not stop at default, which is a label like any case, so d is printed too.

Does [[fallthrough]] change what this program prints, and why would you add it?
Why may a variable declared right after a case label, without braces, cause a compile error?

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JuniorTheoryCommon

What does the program print?

C++

#include <iostream>

int main() {
    int n = 2;
    switch (n) {
        case 1: std::cout << 'a'; break;
        case 2: std::cout << 'b';
        case 3: std::cout << 'c';
        default: std::cout << 'd';
    }
}

bcd

Expecting b: a case label does not end the code above it, so without break execution runs on into case 3 and further.
Expecting bc: falling through does not stop at default, which is a label like any case, so d is printed too.

Does [[fallthrough]] change what this program prints, and why would you add it?
Why may a variable declared right after a case label, without braces, cause a compile error?

Back to questions Open topic in roadmap →Answer it yourself — Quiz