c++ - 获取模板化函数的地址是否应该触发其编译？

Question

我得到了 this question 的官方答复decltype 应该不 触发函数编译。事实上，声明但未定义的函数的 decltype 是合法的。

下一个问题，获取函数的地址是否应该触发函数的编译？取this example :

template <typename T>
void foo(T&& x) { x.func(); }

int main()
{
    auto bar = &foo<int>;
}

我测试过的所有编译器都失败并出现如下错误:

Request for member func in x, which is of non-class type int

但是，如果我只定义 foo 并且不声明它，则代码可以正常编译。有人可以向我提供有关获取函数地址是否需要编译的官方资源吗？

Answer 1

3.2/2:

An expression is potentially evaluated unless it is an unevaluated operand (Clause 5) or a subexpression thereof. ... A non-overloaded function whose name appears as a potentially-evaluated expression or a member of a set of candidate functions, if selected by overload resolution when referred to from a potentially-evaluated expression, is odr-used, unless it is a pure virtual function and its name is not explicitly qualified.

然后 3.2/3:

Every program shall contain exactly one definition of every non-inline function or variable that is odr-used in that program; no diagnostic required. The definition can appear explicitly in the program, it can be found in the standard or a user-defined library, or (when appropriate) it is implicitly defined (see 12.1, 12.4 and 12.8). An inline function shall be defined in every translation unit in which it is odr-used.

函数名称绝对不是未求值的操作数(例如 sizeof、decltype)，并且它出现在表达式中，因此可能会求值。然后第二个需要在每个翻译单元中恰好有一个非内联定义，或相同的内联定义。