为什么一个const数组不能从constexpr函数访问? [英] why is a const array not accessible from a constexpr function?

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问题描述

我有一个名为access的constexpr函数,我想从数组中访问一个元素:

i have a constexpr function named access, and i want to access one element from an array:

char const*const foo="foo";
char const*const bar[10]={"bar"};

constexpr int access(char const* c) { return (foo == c); }     // this is working
constexpr int access(char const* c) { return (bar[0] == c); }  // this isn't
int access(char const* c) { return (bar[0] == c); }            // this is also working

我得到错误:

error: the value of 'al' is not usable in a constant expression

为什么我无法访问其中一个元素?

why can't i access one of the elements from access? or better how do i do it, if it is even possible?

推荐答案

数组需要声明 constexpr ,而不仅仅是 const

The array needs to be declared constexpr, not just const.

constexpr char const* bar[10]={"bar"};

否则,表达式 bar [0] 执行左值到右值转换,以便取消引用数组。除非数组 constexpr ,根据C ++ 11 5.19 / 2,第九个项目符号:

Without that, the expression bar[0] performs an lvalue-to-rvalue conversion in order to dereference the array. This disqualifies it from being a constant expression, unless the array is constexpr, according to C++11 5.19/2, ninth bullet:


一个左值到右值的转换,除非它应用到

an lvalue-to-rvalue conversion unless it is applied to


  • 一个glvalue引用使用constexpr定义的非易失性对象的字面类型

在这里不适用的例外)。

(and a couple of other exceptions which don't apply here).

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