Jump bypasses variable initialization in switch statement

Question

I need an std:: vector<char> or an std:: string in my switch case for some purpose. So, I wrote the following dummy code to see if it works:

#include <iostream>
#include <string>
int main() {
    int choice = 0;

    do {
        std:: cout << "Enter Choice" << std::endl;
        std:: cin >> choice;

        switch(choice) {
            case 1:
                std::cout << "Hi";
                break;

            case 2:
                std::string str;
                std::cin >> str;
                break;

            case 3: //Compilation error, Cannot jump from switch statement to this case label
                std::cout << "World" << std:: endl;
                break;

            default:
                std:: cout << "Whatever" << std:: endl;
        }

    } while(choice != 5);
    return 0;
}

Ok, I somewhat got it that str is an object of std:: string type. So, I am trying to jump through this variable initialization.

Then why a definition of C style string does not cause compilation error:

#include <iostream>
#include <string>
int main() {
    int choice = 0;

    do {
        std:: cout << "Enter Choice" << std::endl;
        std:: cin >> choice;

        switch(choice) {
            case 1:
                std::cout << "Hi";
                break;

            case 2:
                char str[6];
                std::cin >> str;
                break;

            case 3:
                std::cout << "World" << std:: endl;
                break;

            default:
                std:: cout << "Whatever" << std:: endl;
        }

    } while(choice != 5);
    return 0;
}

How can I make the first code to work?

Answer 1

char str[6]; is default-initialized. In case of C arrays of simple values with automatic storage duration ("allocated on stack") it means "no initialization at all", so I guess it's not an error.

If you, however, initialize the array like char str[6] = {}, it will yield an error.

I suggest you put extra curly braces so str is in its own scope and is not visible in further case statements:

#include <iostream>
#include <string>
int main() {
    int choice = 0;

    do {
        std:: cout << "Enter Choice" << std::endl;
        std:: cin >> choice;

        switch(choice) {
            case 1:
                std::cout << "Hi";
                break;

            case 2: {  // Changed here
                std::string str;
                std::cin >> str;
                break;
            }
            case 3: // `str` is not available here, no compilation error
                std::cout << "World" << std:: endl;
                break;

            default:
                std:: cout << "Whatever" << std:: endl;
        }

    } while(choice != 5);
    return 0;
}

Where to put brackets is a styling preference.

Answer 2

Just create a new block for the variable with an extra pair of braces

        case 2:
        { // <-- start new block for str
            std::string str;
            std::cin >> str;
            break;
        } // <-- end of block, str will be destroyed here