comparison between string literal
c++, comparison, literals, string, warnings
Solution
char arrays or char pointers aren't really the same thing as string class objects in C++, so this
if (option == "foo")
Doesn't compare the string `option` to the string literal "foo" it compares the address of `option` with the address of the string literal "foo". You need to use one of the many string comparison functions if you want to know if the option is the same as "foo". `strcmp` is the obvious way to do this, or you can use `std::string` instead of `char*`
Problem
This very simple code: ``` #include <iostream> using namespace std; void exec(char* option) { cout << "option is " << option << endl; if (option == "foo") cout << "option foo"; else if (option == "bar") cout << "opzion bar"; else cout << "???"; cout << endl; } int main() { char opt[] = "foo"; exec(opt); return 0; } ``` generate two warning: comparison with string literal results in unspecified behaviour. Can you explain why exactly this code doesn't work, but if I change ``` char opt[] ``` to ``` char *opt ``` it works, but generates the warning? Is it related to the \0 termination? What is the difference between the two declaration of opt? What if I use const qualifier? The solution is to use std::string?