Use of constexpr function before definition fails

Question

I'm having some trouble with constexpr. The book C++ Primer shows a line of code:

  constexpr int sz = size(); // only size() is a constexpr function                              // this code is right 

However the book doesn't give a specific example. So I try the following code by myself:

#include <iostream> constexpr int fun(); int main() {     constexpr int f = fun();     std::cout << f << std::endl; } constexpr int fun() {     return 3; } 

But my compiler said fun() is undefined.

If I change constexpr into const, it works well, and if I change my code to define the constexpr function before use:

#include <iostream> constexpr int fun() {     return 3; } int main() {     constexpr int f = fun();     std::cout << f << std::endl; } 

It also works well. Can someone tell me why?

Answer 1

A constexpr function does NOT have to be defined before its first use, however the result of any call made prior to definition is not a constant expression.

Source: C++ Standard draft n4296, section 5.20:

A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

• this, except in a constexpr function or a constexpr constructor that is being evaluated as part of e;
• an invocation of a function other than a constexpr constructor for a literal class, a constexpr function, or an implicit invocation of a trivial destructor [ Note: Overload resolution is applied as usual — end note ];
• an invocation of an undefined constexpr function or an undefined constexpr constructor;
• ...

version from draft 3485 (section 5.19):

A conditional-expression is a core constant expression unless it involves one of the following as a potentially evaluated subexpression, but subexpressions of logical AND, logical OR, and conditional operations that are not evaluated are not considered [ Note: An overloaded operator invokes a function. — end note ]:

• this [ Note: when evaluating a constant expression, function invocation substitution replaces each occurrence of this in a constexpr member function with a pointer to the class object. — end note ];
• an invocation of a function other than a constexpr constructor for a literal class or a constexpr function [ Note: Overload resolution is applied as usual — end note ];
• an invocation of an undefined constexpr function or an undefined constexpr constructor
• ...

The example int x2 = s. t(); in n2235 actually became valid due to the changes made prior to Standardization. However, constexpr int x2 = s. t(); remains an error.