为什么C ++ 11不支持指定的初始化列表为C99？ [英] Why does C++11 not support designated initializer list as C99?

问题描述

struct Person
{   
    int height;
    int weight;
    int age;
};

int main()
{
    Person p { .age = 18 };
}

上面的code为C99合法的，但在C ++ 11不合法的。

The code above is legal in C99 but not legal in C++11.

那是C ++ 11不支持这样一个方便的功能的理由？

推荐答案

C ++有构造函数。如果是有意义的初始化只是一个成员，那么，可以通过实施适当的构造函数在程序pssed前$ P $。这是哪门子的抽象C ++的提升。

C++ has constructors. If it makes sense to initialize just one member then that can be expressed in the program by implementing an appropriate constructor. This is the sort of abstraction C++ promotes.

在另一方面，指定初始化功能更多的是暴露，使成员容易直接访问客户端code。这导致了事情就像一个人的年龄18（岁？），但与零的身高和体重。

On the other hand the designated initializers feature is more about exposing and making members easy to access directly in client code. This leads to things like having a person of age 18 (years?) but with height and weight of zero.

在换句话说，指定的初始化程序支持的编程风格，其中内部暴露，并在客户端被赋予的灵活性，以决定如何使用类型。

In other words, designated initializers support a programming style where internals are exposed, and the client is given flexibility to decide how they want to use the type.

C ++更感兴趣的是把灵活的一边在的设计师的一个类型，而不是，因此设计人员可以很容易地正确使用类型和困难使用不正确。把设计师的类型如何被初始化控制是这一部分：设计人员决定构造函数，在类初始化等

C++ is more interested in putting the flexibility on the side of the designer of a type instead, so designers can make it easy to use a type correctly and difficult to use incorrectly. Putting the designer in control of how a type can be initialized is part of this: the designer determines constructors, in-class initializers, etc.

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