How do I ignore a particular gcc warning that's [enabled by default]?

I have the following program that prints green text to the terminal:

#include <iostream> #include <string> //returns a colored string for terminal output streams std::string colorize_forground(std::string const& message, int const& background) { return std::string("\e[38;5;" + std::to_string(background) + "m" + message + "\x1b[0m"); } int main() { std::cout << colorize_forground("hello world in green", 106) << '\n'; } 

However, when I compile the program with the following warning flag,

g++ -std=c++1y -pedantic -o main prob.cpp

I get this warning message:

main.cpp: In function ‘std::string colorize_forground(const string&, const int&)’: main.cpp:6:21: warning: non-ISO-standard escape sequence, '\e' [enabled by default] return std::string("\e[38;5;" + std::to_string(background) + "m" + message + "\x1b[0m"); 

How do I continue using -pedantic, but ignore the warning for this particular function?

I've been trying to use gcc's Diagnostic Pragmas to ignore this escape sequence warning. I wrapped the function the following way, but it still elicits a warning.

#pragma GCC diagnostic push #pragma GCC diagnostic ignored "-pedantic" std::string colorize_forground(std::string const& message, int const& background) { return std::string("\e[38;5;" + std::to_string(background) + "m" + message + "\x1b[0m"); } #pragma GCC diagnostic pop