c++ - 关于constexpr函数的一个问题

Question

先看代码

#include <iostream>
#include <cstdlib>
#include <cstdio>
//#include "test.h"

using namespace std;

constexpr int screen(int x)
{
   return x; 
}


int main()
{   
    int x = 0;
    int z =  screen(x);
    cout << z<<endl;
    return 0;
}

  在C++ Primer一书中说到constexpr函数定义的约定：函数的返回类型以及所有的形参类型必须是字面值类型（即编译过程就能得到结果的类型）。但是同时又说constexpr函数不一定返回常量表达式。感觉前后有一些矛盾，就像上面的代码一样，通过 g++ test.cpp -std=c00x 的编译运行都是没有问题的，这里应该怎么理解？

Answer 1

看看这个:A constexpr function is not required to return a constant expression?

Answer 2

C++ standard document n45675.20 section talks about this

int main()
{   
    int x = 0;
    int z =  screen(x); //因为z是非constexpr的，不要求screen是constexpr的
    cout << z<<endl;
    return 0;
}

int main()
{   
    int x = 0;
    constexpr int z =  screen(x); //错误！！ x的lifetime开始于screen(x)的外部
    cout << z<<endl;
    return 0;
}

int main()
{  
    constexpr int z =  screen(0); //screen(0)是常量表达式, 对于screen内部，x的lifetime开始于函数内部
    cout << z<<endl;
    return 0;
}

Answer 3

You can think of it this way: C++ does not require that the constexpr function must return a constant expression (take the screen constexpr function in the question as an example)

• If you are in a context that does not require a constant expression, such as: int z = screen(x); You do not need to return a constant expression. At this time, the compiler will not check whether the result of the function will return a constant expression.

• If it is in a context that requires a constant expression, such as: constexpr int z = screen(x); Then the constexpr function must be able to return a constant expression. At this time, the compiler will check whether the result returned by the function is a constant expression, and if not, an error will be reported.