无法在函数定义的类外声明符中完全限定类名 [英] Impossible to fully qualify class-name in out-of-class declarator of function definition

问题描述

此程序会导致不希望的解析贪婪的死角:

This program results in an undesired parsing greediness dead-end:

struct float4x4 {};
class C
{
    float4x4 M();
};

float4x4 ::C::M()
{
    return float4x4{};
}

:8:1:错误:'float4x4'中没有名为'C'的成员;你的意思仅仅是"C"吗?
float4x4 :: C :: M()
^ ~~~~~~~~~~~

:8:1: error: no member named 'C' in 'float4x4'; did you mean simply 'C'?
float4x4 ::C::M()
^~~~~~~~~~~~

可以使用尾随返回类型固定"哪些内容:

Which can be 'fixed' using trailing return type:

auto ::C::M() -> float4x4
{}

现在一切都好.

所以我认为在使用heading-return-type声明符语法时，我们不能完全限定类名吗?

So I take it we can't fully qualify the class-name when using heading-return-type declarator syntax?

推荐答案

您可以使用方括号消除歧义:

You can put brackets to disambiguate:

float4x4 (::C::M)()
{
    return float4x4{};
}

虽然我使用gcc和clang(均为-pedantic)进行了测试，但我不能真正告诉您什么规则可以使之正常运行，尽管并非没有括号.我更喜欢尾随返回类型.

I cannot really tell you what rule makes this ok, while it is not without the brackets, though I tested with gcc and clang (both -pedantic). I would prefer the trailing return type.

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