CODEWITHSUNDEEP
c++constexprconstexpr-function
icecity96
icecity96

Reputation: 1227

Use of constexpr function before definition fails

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 clang 3.4.1 says:

undefined function 'fun' cannot be used in a constant expression

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?

Upvotes: 31

Views: 3521

Answers (2)

Ben Voigt
Ben Voigt

Reputation: 283624

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.

Upvotes: 24

kvorobiev
kvorobiev

Reputation: 5070

A constant expression function must be defined before its first use. See this paper, end of section 4.1.

Upvotes: 3

Related Questions