为什么不能在类范围内推导我的类静态自动函数的类型？ [英] Why can't the type of my class-static auto function be deduced within the class scope?

问题描述

我正在尝试获取 auto 函数的返回类型。 此方法：

I'm trying to get the return type of an auto function. This works:

auto foo(int bar)
{
    return 0;
}

typedef std::result_of<decltype(foo)> foo_t;

很好，接下来是下一步：获取 static的返回类型类范围内的auto 函数。 这也有效：

Great, here's the next step then: getting the return type of a static auto function in a class scope. This also works:

struct Foo
{
    static auto foo(int bar)
    {
        return 0;
    }
};

typedef std::result_of<decltype(Foo::foo)> foo_t;

但是这不起作用：

struct Foo
{
    static auto foo(int bar)
    {
        return 0;
    }

    typedef std::result_of<decltype(Foo::foo)> foo_t;
};

GCC表示错误：在扣除前，使用'static auto Foo :: foo（int）' Clang说道：'auto'，在定义之前不能使用带有推导返回类型的函数'foo'。为什么？

GCC says "error: use of 'static auto Foo::foo(int)' before deduction of 'auto'", Clang says "function 'foo' with deduced return type cannot be used before it is defined". Why?

推荐答案

虽然您编写代码的方式使之成为可能，但<$ c $的类内定义c> foo（）仅在完全定义该类之后才能进行处理。好像您是这样写的：

While the way you have written the code makes it appear possible, the in-class definition of foo() can only be processed after the class is fully defined. It is as if you wrote this:

struct Foo
{
    static auto foo(int bar);

    typedef std::result_of<decltype(Foo::foo)> foo_t;
};

auto Foo::foo(int bar)
{
    return 0;
}

foo（）使用类Foo 中定义的类型，包括 foo_t ，这些类型可以是循环的。因此，类Foo 的定义不允许使用其成员函数的定义-仅使用其声明。

The definition of foo() is allowed to use types defined in class Foo, including foo_t, which would be circular. Therefore, the definition of class Foo is not allowed to use the definition of its member functions--only their declarations.

换句话说，您假设代码是从上到下完全评估的。不是。

In other words, you assume the code is fully evaluated from the top to the bottom. It is not.

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