初始化字符串时额外花括号 [英] Extra curly braces while initializing a string

问题描述

根据问题此代码的含义是什么，为什么它会工作？我想问下面代码中 s1 和 s2 之间的区别：

  int main（）{
 const char * s1 = {Hello}; //奇怪，但工作如下
 const char * s2 =Hello; //普通case 
 return 0; 
} 
  

为什么要使用额外花括号？任何对C ++标准的引用将是有用的。

解决方案

在C ++ 98（和C ++ 03） ;在第8.5节中：

14 - 如果 T 是标量类型，声明形式
T x = {a};
相当于
T x = a;




在C ++ 11中，这由列表初始化（8.5.4p3）覆盖：，那么p>

引用类型是引用相关的E，对象或引用从该元素初始化[...]

认为这是与使用大括号初始化标量相同的问题。

According to question What does this code mean and why does it work? I want to ask what difference between s1 and s2 in the code below:

int main() {
    const char* s1 = { "Hello" }; // strange but work as followed
    const char* s2 = "Hello"; // ordinary case 
    return 0;
}

Why extra curly braces are permitted? Any reference to C++ standard will be useful.

解决方案

In C++98 (and C++03) this is pretty simple; in clause 8.5:

14 - If T is a scalar type, then a declaration of the form T x = { a }; is equivalent to T x = a;

In C++11 this is covered by list-initialization (8.5.4p3):

[...] if the initializer list has a single element of type E and either T is not a reference type or its referenced type is reference-related to E, the object or reference is initialized from that element [...]

I think this is the same question as Initializing scalars with braces.

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