为什么ISO C ++标准禁止成员的初始化？ [英] Why the ISO C++ standard forbids the initialization for members?

问题描述

为什么我不能这样做

class A {
  public:

    int x = 10;

  ...
};

我必须这样做吗？

class A {
  public:

    int x;

  A(){
    x = 10;
    ...
  }

  ...
};

这是因为C ++试图比C语言更安全吗？还有其他原因吗？

It's because C++ is trying to be more type safe than languages like C ? There are other reasons ?

推荐答案

这与类型安全无关，两个例子都是安全的。

This has nothing to do with type safety, both examples are as safe.

创建语言时，您需要定义允许的内容和不允许的内容。因为写一个黑名单是一个永无止境的经历，一种语言通常是以白名单的方式写的，增加了越来越多的可能的东西。

When creating a language you need to define what is allowed and what is not. Since writing a blacklist would be a never-ending experience, a language is usually written in a white-list fashion, adding more and more possible things.

但是，不允许事情没有明确的权衡后果。每当你想允许新的东西，你需要检查：

However, one should not allow things without clearly weighing the consequences. Whenever you want to allow something new, you need to check:

• 易于实现


• 与其他现有功能的互动和已知的可能扩展

此外，这也意味着有更多的学习希望使用语言。

Furthermore, it also means that there is more to learn for those wishing to use the language.

但是，这里要求的是比较容易的。它可以看作是对等体，对于类，函数默认参数。这在C ++ 11中被采用，因为它是在写多个构造函数时保证默认值的一致性的最简单的方法。

What you are asking for here is, however, relatively easy. It can be seen as the counterpart, for classes, to functions default arguments. This has been adopted in C++11 because it was the simplest way to guarantee a coherence of the default values when writing multiple constructors.

当使用C ++ 11，我建议以这种方式初始化所有内置类型（如果只有 0 ），很像我建议在声明局部变量时初始化它们：这种方式你不能忘记初始化他们在一个构造函数中。

Personally, when using C++11, I recommend initializing all built-in types this way (if only to 0), much like I recommend initializing them when declaring local variables: this way you cannot forget to initialize them in one of the constructors.

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