定义之前使用constexpr函数失败 [英] Use of constexpr function before definition fails

问题描述

我遇到了一些麻烦与 constexpr 。书 C ++ Primer 显示一行代码：

  constexpr int sz = size //只有size（）是一个constexpr函数
 //这段代码是正确的
  

这本书没有给出具体的例子。所以我自己尝试下面的代码：

  #include< iostream> 
 constexpr int fun（）; 
 int main（）
 {
 constexpr int f = fun（）; 
 std :: cout<< f<< std :: endl; 
} 
 constexpr int fun（）
 {
 return 3;但是我的编译器说 fun（） 
 
 
 
 
 
 
 
 
 
 
如果我将 constexpr 更改为 const ，则code未定义
 
 
 < c $ c>，它工作得很好，如果我改变我的代码在使用之前定义constexpr函数：
  #include< ; iostream> 
 constexpr int fun（）
 {
 return 3; 
} 
 int main（）
 {
 constexpr int f = fun（）; 
 std :: cout<< f<< std :: endl; 
} 
  
有人可以告诉我为什么？
解决方案
一个 constexpr  
 
 
 
源代码：C ++标准草案n4296，第5.20节：
 / p> 
 
 
 
 
条件表达式  e 是核心常量表达式，除非遵循抽象机器的规则的 e 的求值将评估以下表达式之一：
 
 
 
 
 
  this ，除了 constexpr 作为 e ; 
 
 
的一部分进行评估的 constexpr 构造函数对于字面类， constexpr 函数的一个 constexpr 构造函数，或者隐式调用析构函数[注意：过载解析应用为
 usual  -  end note]; 
 
 
 调用未定义的 constexpr  function 或未定义的 constexpr 构造函数; 
 
 
 ... 
 
 
 
 
版本从草案3485（第5.19节）：
 
 <除非涉及以下之一作为潜在求值的子表达式，否则条件表达式是核心常量表达式，但是逻辑AND，逻辑OR和条件操作的子表达式不会被考虑[注意：重载的操作符调用一个函数。 -  end note]：
 
 
 
 
 
  this  [注意：函数调用替换在 constexpr 成员函数中用指向类对象的指针替换每次出现的 this 。 -  end note]; 
 
 
调用函数而不是 constexpr 构造函数用于文字类或 constexpr  function [注意：过载分辨率像平常一样应用 -  end note]; 
 
 
 调用未定义的 constexpr  function 或未定义的 constexpr 构造函数
 
 
 ... 
 
 
 
 
示例 int x2 = s。 t（）; 在n2235实际上变得有效，由于在标准化之前做出的更改。但是， constexpr int x2 = s。 t（）; 仍然是错误。
 
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?
解决方案
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.

                        
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