自动与字符串文字 [英] auto with string literals

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!\n";

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

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

为什么 x

一个窄字符串文字的类型为 n const char [2.14.5 String Literals [lex.string]§8]

A narrow string literal has type "array of n const char" [2.14.5 String Literals [lex.string] §8]

推荐答案

特征 auto 基于模板参数扣除和模板参数扣除行为相同，特别是根据§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来代替A来进行类型推导。
  
  

  如果你希望表达式 x 的类型是一个数组类型， $ c>& 之后：

  If you want the type of the expression x to be an array type, just add & after auto:

  auto& x = "Hello world!";
  

  然后，将推导出 auto 占位符要 const char [13] 。这也类似于以引用作为参数的函数模板。为了避免混淆：声明的x类型将是引用 -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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