问题描述

类内初始化器(C++11 特性)必须用 curl 括起来大括号或跟随一个 = 符号.它们不能在括号内指定.

In-class initializers (C++11 feature) must be enclosed in curly braces or follow a = sign. They may not be specified inside parenthesis.

这是什么原因?

推荐答案

我对此并不是 100% 肯定，但这可能是为了防止语法歧义.例如，考虑以下类:

I am not 100% positive about this, but this might be to prevent a syntax ambiguity. For example, consider the following class:

class BadTimes {
    struct Overloaded;
    int Overloaded;            // Legal, but a very strange idea.

    int confusing(Overloaded); // <-- This line
};

指示的行是什么意思?正如所写，这是一个名为 confusing 的成员函数的声明，它接受一个 Overloaded 类型的对象作为参数(其名称未在函数声明中指定)和返回一个 int.如果 C++11 允许初始化器使用括号，这将是模棱两可的，因为它也可能是一个名为 confusing 的 int 类型成员的定义，该成员被初始化到数据成员 Overloaded 的值.(这与 Most Vexing Parse 的当前问题有关.)

What does the indicated line mean? As written, this is a declaration of a member function named confusing that accepts as a parameter an object of type Overloaded (whose name isn't specified in the function declaration) and returns an int. If C++11 were to allow initializers to use parentheses, this would be ambiguous, because it could also be a definition of a member of type int named confusing that is initialized to the value of the data member Overloaded. (This is related to the current issue with the Most Vexing Parse.)

通过要求使用大括号，消除了这种歧义:

By requiring curly braces, this ambiguity is removed:

class BadTimes {
    struct Overloaded;
    int Overloaded;            // Legal, but a very strange idea.

    int confusing{Overloaded}; // <-- This line
};

现在，很明显 confusing 实际上是一个 int 初始化为 Overloaded 的值，因为没有办法将它读作函数声明.

Now, it's clear that confusing is actually an int initialized to the value of Overloaded, because there's no way to read it as a function declaration.

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