c++ - 不允许从函数返回函数。我怎么能？

Question

8.3.5/8 函数 [dcl.fct] 说

[...] Functions shall not have a return type of type array or function, although they may have a return type of type pointer or reference to such things. [...]

为什么规则如此明确？是否有一些语法甚至允许返回函数而不是函数指针？

我是否误解了这句话？

typedef void (*fp)();

void foo(){}
fp goo()
{
    return foo; //automatically converted to function pointer
}

Answer 1

这是一个非常人为的函数尝试返回函数的例子:

void foo() { }

template<typename T>
T f() { return foo; }

int main(){
    f<decltype(foo)>();
}

这是我从 Clang 3.2 得到的错误:

Compilation finished with errors:
source.cpp:7:5: error: no matching function for call to 'f'
    f<decltype(foo)>();
    ^~~~~~~~~~~~~~~~
source.cpp:4:3: note: candidate template ignored: substitution failure 
[with T = void ()]: function cannot return function type 'void ()'
T f() { return foo; }
~ ^
1 error generated.