Returning a locally created const char*

Question

#include <iostream>


const char* fun()
{
    const char* x = "abc";
    std::cout << "x = " << x << "\n";
    return x;
}


int main(int arc, char** argv)
{
    const char* y = fun();
    std::cout << "y = " << y << "\n";
    return 0;
}

Running this on my machine gives:

x = abc

y = abc

In fun(), x (a local variable) is assigned the address of a string literal created locally, yet when the function returns, the data pointed to by y is the same as that pointed to by x even though x is out of scope.

Can someone explain in detail what is happening here?

Answer 1

This is well-formed, the returned pointer is valid and not dangled; because the string literal (i.e. "abc") has static storage duration and exists in the whole life of the program.

String literals have static storage duration, and thus exist in memory for the life of the program.

As you said when the function returns the local variable x gets destroyed, but the string literal pointed to by it doesn't.