声明一个类的 const 实例

2022-01-23 00:00:00 constants c++

假设我有一个定义如下的类:

Let's say I have a class defined as follows:

class foo{};

现在，这是完全可以接受的；

now, this is perfectly acceptable;

foo f;

这怎么会是编译器错误?(未初始化的 const ‘f’)

how come this is a compiler error? (uninitialized const ‘f’)

const foo f;

为什么我们必须这样做?

Why do we have to do this?

const foo f = foo();

我知道为什么我们不能这样做..

I know why we can't do this..

const foo f(); // though it compiles..

有趣的是，以下是有效的:

Interestingly, the following is valid:

const std::string f;

那么 foo 中缺少什么?

我知道那里有三个问题，而且形式不好，但我希望有人可以在一个答案中为我解决这个问题.

I realize that there are three questions there and it's bad form, but I'm hoping someone can clear this up for me in one answer.

如果它很愚蠢，请随时关闭它......

please feel free to close it if it's stupid...

推荐答案

你的类是一个 POD(本质上是因为它不提供默认构造函数).POD 变量在声明时不初始化.也就是这个:

Your class is a POD (essentially because it doesn’t provide a default constructor). POD variables are not initialized upon declaration. That is, this:

foo x;

不会将 x 初始化为有意义的值.这必须单独进行.现在，当您将其声明为 const 时，这可能永远不会发生，因为您不能再分配或更改 x.

does not initialize x to a meaningful value. This has to be done separately. Now, when you declare it as const, this may never happen because you cannot assign to or change x any more.

考虑与 int 的等价性:

int x; // legal
const int y; // illegal

正如您所注意到的，使用 std::string 而不是 foo 可以编译.那是因为 std::string 不是 POD.一个简单的解决方案是为 foo 提供一个默认构造函数:

As you have noticed, using std::string instead of foo compiles. That’s because std::string is not a POD. A simple solution to your dilemma is to provide a default constructor for foo:

class foo {
public:
    foo() { }
};

现在你的 const foo x; 代码编译好了.

Now your const foo x; code compiles.

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