Why is my function returning garbage when it should return a char?

Question

I\'m a newbie in C++ learning the language and playing around. I wrote a piece of code which behavior I don\'t understand. Could someone explain why the code below prints ou

Answer 1

It's because you return a pointer to a local variable, a local variable that goes out of scope when the function returns.

You are already using std::string for the argument, use it instead of the array and the return pointer.

Answer 2

Your function is returning garbage because you're returning the address of a local variable which goes out of scope after your function returns. It should probably look like this:

char* str2char(const std::string &str)
{
    char *const cset = new char[str.size() + 1]; // +1 for the null character
    strcpy(cset, str.c_str());
    return cset;
}

You will need to delete your variable r by doing delete[] r;. Ideally though you wouldn't be using raw pointers, and you would use std::string for everything, or wrap the char * in a std::unique_ptr.

Answer 3

If your aim is to pass the content of a std::string to a function modifying the content of a char*:

#include <iostream>
#include <vector>

void f(char* s) {
    s[0] = 'H';
}

std::vector<char> to_vector(const std::string& s) {
    return std::vector<char>(s.c_str(), s.c_str() + s.size() + 1);
}

int main(void)
{
    std::string s = "_ello";
    std::vector<char> t = to_vector(s);
    f(t.data());
    std::cout <<  t.data() << std::endl;
}