Sepples Magic


#include <string>
#include <vector>
#include <iostream>

std::vector<std::string> names;

void addNames()
{
    names.push_back(std::string("Foo"));
    names.push_back(std::string("Bar"));
}

int main(int argc, char **argv)
{
    addNames();
   
    for (int i = 0, size = names.size(); i < size; ++i)
    {
        std::cout << names[i] << std::endl;
    }

    return 0;
}

Output:

Foo
Bar

Okay so what I'm trying to understand is since I'm pushing anonymous stack string objects onto the vector, I'm assuming that the strings will also have their destructors called (because local variables). This should also destroy the internal char sequence.

Yet the strings remain intact when printing the vector (so somehow its being copied?). So /prog/ please show me your sepples mastery and explain the magic here.

>>1
Okay so what I'm trying to understand is since I'm pushing anonymous stack string objects onto the vector,

You are pushing copies of them actually. Which use copies of the string constants themselves. In other words: you have a char* constant "Foo" allocated in the program's const data area, a temporary stack-allocated string object and a heap region containing the copy of "Foo", and an element of a vector<string> holding a pointer to another heap region containing the copy of "Foo". The destruction of the temporary doesn't affect either of the others.

You now also see how C++ allows utilizing every last bit of performance of the hardware!