为什么在使用decltype时，编译器为什么不推断成员的类型? [英] Why doesn't the compiler infer the type of member when using decltype?

问题描述

我只是在自己的代码中注意到了这种行为，所以这是一个天真的问题:

I just noticed this behavior in own code, so here is the naive question:

此:

struct A
{
    int get()
    {
        return a;
    }
    int a=1;
};   
int main() {}

当然可以编译，尽管当成员数据的声明位于函数定义的之后.

Compiles of course fine, although when the declaration of member data lies after the function definition.

但是我不明白为什么这样:

But then I don't understand why this:

struct A
{
    auto get() -> decltype(a)
    {
        return a;
    }
    int a=1;
};  

不编译(*).我必须写这个:

does not compile(*). I have to write this:

struct A
{
    int a=1;
    auto get() -> decltype(a)
    {
        return a;
    }
};  

是否有任何与语言相关的原因导致它不正常，或者仅仅是编译器没有实现?无论类成员的顺序如何，我都希望具有相同的行为.

Is there any language-related reasons why it's not ok, or is it just that the compilers have not implemented that ? I would expect to have the same behavior regardless of the order of the class members.

(*)通过Ideone.com在gcc 6.3上进行了测试

(*) tested with gcc 6.3 through Ideone.com

推荐答案

在您的第一个示例中，声明了 A :: get ，然后声明了 A :: a ，直到那时(因为完全声明了 A )，才定义了 A :: get .此时，在 A :: get 中，编译器知道了 A :: a .

In your first example, A::get is declared, then A::a is declared, and only then (because A is fully declared), A::get is defined. At this point, in A::get, the compiler knows about A::a.

考虑以下等效形式:

struct A
{
    int get();
    int a=1;
};

inline int A::get()
{
    return a;
}

这对于您的第二个示例来说是无法编译的:

This would not be compilable for you second example:

struct A
{
    auto get() -> decltype(a); // What is "a"?
    int a=1;
};

inline auto A::get() -> decltype(A::a)
{
    return a;
}

声明类型时必须遵循相同的声明顺序"，例如:

The same "order of declaration" has to be respected when you declare types, for example:

struct A
{
    using IntType = int;
    auto get() -> IntType;
    int a=1;
};

你不能这样写:

struct A
{
    using IntType = decltype(a);
    auto get() -> IntType;
    int a=1;
};

还请注意，这不仅限于返回类型:参数类型也是函数声明的一部分.因此，这也不会编译:

Also note that this is not limited to return types: parameter types are also part of a function declaration. So this doesn't compile either:

struct A
{
    void f(decltype(a));
    int a=1;
};

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