Does constexpr imply inline?

问题:

Consider the following inlined function :

// Inline specifier version
#include<iostream>
#include<cstdlib>

inline int f(const int x);

inline int f(const int x)
{
    return 2*x;
}

int main(int argc, char* argv[])
{
    return f(std::atoi(argv[1]));
}

and the constexpr equivalent version :

// Constexpr specifier version
#include<iostream>
#include<cstdlib>

constexpr int f(const int x);

constexpr int f(const int x)
{
    return 2*x;
}

int main(int argc, char* argv[])
{
    return f(std::atoi(argv[1]));
}

My question is : does the constexpr specifier imply the inline specifier in the sense that if a non-constant argument is passed to a constexpr function, the compiler will try to inline the function as if the inline specifier was put in its declaration ?

Does the C++11 standard guarantee that ?

回答1:

Yes ([dcl.constexpr], §7.1.5/2 in the C++11 standard): "constexpr functions and constexpr constructors are implicitly inline (7.1.2)."

Note, however, that the inline specifier really has very little (if any) effect upon whether a compiler is likely to expand a function inline or not. It does, however, affect the one definition rule, and from that perspective, the compiler is required to follow the same rules for a constexpr function as an inline function.

I should also add that regardless of constexpr implying inline, the rules for constexpr functions in C++11 required them to be simple enough that they were often good candidates for inline expansion (the primary exception being those that are recursive). Since then, however, the rules have gotten progressively looser, so constexpr can be applied to substantially larger, more complex functions.