当将重载函数作为参数时，模板参数推导如何工作? [英] How does template argument deduction work when an overloaded function is involved as an argument?

问题描述

这是更复杂的问题 -is-an-overloaded-function>当参数是重载函数时，重载解析如何工作?

This is the more sophisticated question mentioned in How does overload resolution work when an argument is an overloaded function?

下面的代码编译没有任何问题:

void foo() {}
void foo(int) {}
void foo(double) {}
void foo(int, double) {}

// Uncommenting below line break compilation
//template<class T> void foo(T) {}

template<class X, class Y> void bar(void (*f)(X, Y))
{
    f(X(), Y());
}

int main()
{
    bar(foo);
}

对于模板自变量的推论似乎不是一项艰巨的任务-只有一个函数foo()接受两个自变量.但是，取消注释foo()的模板重载(仍然只有一个参数)不会毫无原因地中断编译.使用gcc 5.x/6.x和clang 3.9编译都会失败.

It doesn't appear a challenging task for template argument deduction - there is only one function foo() that accepts two arguments. However, uncommenting the template overload of foo() (which still has just a single parameter) breaks compilation for no obvious reason. Compilation fails both with gcc 5.x/6.x and clang 3.9.

可以通过重载解析/模板参数推导的规则来解释它，还是应该将其视为那些编译器中的缺陷?

Can it be explained by the rules of overload resolution/template argument deduction or it should be qualified as a defect in those compilers?

推荐答案

如所链接问题的答案中所述:

As noted in the answer to your linked question:

[temp.deduct.call]/6:当P是函数类型时，指向函数类型的指针或指向成员函数类型的指针:

[temp.deduct.call]/6:When P is a function type, pointer to function type, or pointer to member function type:

-如果参数是包含一个或多个函数模板的重载集，则将参数视为 作为非推论上下文.

— If the argument is an overload set containing one or more function templates, the parameter is treated as a non-deduced context.

由于重载集包含一个功能模板，因此该参数被视为非推导上下文.这会导致模板参数推导失败:

Since the overload set contains a function template, the parameter is treated as a non-deduced context. This causes template argument deduction to fail:

[temp.deduct.type]/4: [...]如果仅在非推导中使用模板参数 上下文且未明确指定，模板参数推导失败.

[temp.deduct.type]/4: [...]If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

这个失败的推论给你你的错误.请注意，如果您明确指定参数，则代码将成功编译:

And this failed deduction gives you your error. Note that if you explicitly specify the arguments, the code compiles successfully:

bar<int,double>(foo);

实时演示

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