带有字符串文字的自动 | 带有字符串文字的自动

本文介绍了带有字符串文字的自动的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

#include <iostream>
#include <typeinfo>

int main()
{
    const char a[] = "hello world";
    const char * p = "hello world";
    auto x = "hello world";

    if (typeid(x) == typeid(a))
        std::cout << "It's an array!
";

    else if (typeid(x) == typeid(p))
        std::cout << "It's a pointer!
";   // this is printed

    else
        std::cout << "It's Superman!
";
}

当字符串文字实际上是数组时，为什么将 x 推导出为指针?

Why is x deduced to be a pointer when string literals are actually arrays?

窄字符串文字的类型为n const char 数组"[2.14.5 字符串文字 [lex.string] §8]

推荐答案

auto 特性基于模板参数推导和模板参数推导行为相同，具体根据 §14.8.2.1/2(C++11 标准):

The feature auto is based on template argument deduction and template argument deduction behaves the same, specifically according to §14.8.2.1/2 (C++11 standard):

如果 P 不是引用类型
- 如果 A 是数组类型，则使用数组到指针转换产生的指针类型代替 A 进行类型推导
如果你想让表达式x的类型为数组类型，只需在auto后添加&:
If you want the type of the expression x to be an array type, just add & after auto:
```
auto& x = "Hello world!";
```
然后，auto 占位符将被推导出为 const char[13].这也类似于将引用作为参数的函数模板.只是为了避免混淆:x 的声明类型将是reference-to-array.
Then, the auto placeholder will be deduced to be const char[13]. This is also similar to function templates taking a reference as parameter. Just to avoid any confusion: The declared type of x will be reference-to-array.

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