Question

Possible Duplicate:
Can a local variable's memory be accessed outside its scope?
why does this function return garbage value

Why does this simple code return garbage?

char *output()
{
    char o[2] = "A";
    return o;
}

int main()
{
    std::cout << output(); 
}

Answer 1

Because you return a pointer to invalid memory – o is destroyed when output returns.

You have several options:

• allocate memory dynamically (using malloc), copy "A" into this memory and return its address
• directly return string literal: return "A";

P.S. Of course, you may use std::string and you will not have this issue. Or use in/out param, instead of return.