constexpr function returning string literal

Question

A function returning a copy of an integer literal

int number()
{ return 1; }

can be easily converted to a plain compile-time expression using the keyword constexpr.

constexpr int number()
{ return 1; }

However, I'm getting confused when it comes to string literals. The usual method is returning a pointer to const char that points to the string literal,

const char* hello()
{ return "hello world"; }

but I think that merely changing "const" to constexpr is not what I want (as a bonus, it also produces the compiler warning deprecated conversion from string constant to 'char*' using gcc 4.7.1)

constexpr char* hello()
{ return "hello world"; }

Is there a way to implement hello() in such a way that the call is substituted with a constant expression in the example below?

int main()
{
    std::cout << hello() << "\n";
    return 0;
}

Shafik Yaghmour Shafik Yaghmour · Accepted Answer · 2014-03-21T02:55:05

const and constexpr are not interchangeable, in your case you do not want to drop the const but you want to add constexpr like so:

constexpr const char* hello()
{
  return "hello world";
}

The warning you receive when you drop const, is because a string literal is an array of n const char and so a pointer to a string literal should be a *const char ** but in C a string literal is an array of char even though it is undefined behavior to attempt to modify them it was kept around for backwards compatibility but is depreciated so it should be avoided.

constexpr function returning string literal

2 Answers