为什么在其他函数内部声明的函数不参与依赖参数的查找? [英] Why doesn't function declared inside other function participate in argument dependent lookup?

问题描述

考虑一个简单的示例:

template <class T>
struct tag { };

int main() {
    auto foo = [](auto x) -> decltype(bar(x)) { return {}; };
    tag<int> bar(tag<int>);
    bar(tag<int>{}); // <- compiles OK
    foo(tag<int>{}); // 'bar' was not declared in this scope ?!
}

tag<int> bar(tag<int>) { return {}; }

[gcc] 和

Both [gcc] and [clang] refuses to compile the code. Is this code ill-formed in some way?

推荐答案

来自不合格的查找规则([basic.lookup.unqual]):

From unqualified lookup rules ([basic.lookup.unqual]):

对于类X的成员，成员函数体[...]中使用的名称应在其中的一个中声明. 以下方式
—如果X是本地类或本地类的嵌套类，则在块中定义类X之前 包含类X

For the members of a class X, a name used in a member function body, [...], shall be declared in one of the following ways
— if X is a local class or is a nested class of a local class, before the definition of class X in a block enclosing the definition of class X

您的通用lambda是main中的局部类，因此要使用bar，名称bar必须事先出现在声明中.

Your generic lambda is a local class within main, so to use bar, the name bar must appear in a declaration beforehand.

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