SFINAE:检测类是否具有自由功能 [英] SFINAE: detect if class has free function

问题描述

有没有一种使用SFINAE的方法来检测给定类的自由函数是否被重载了?

Is there a way, using SFINAE, to detect whether a free function is overloaded for a given class?

基本上，我有以下解决方案:

Basically, I’ve got the following solution:

struct has_no_f { };

struct has_f { };

void f(has_f const& x) { }

template <typename T>
enable_if<has_function<T, f>::value, int>::type call(T const&) {
    std::cout << "has f" << std::endl;
}

template <typename T>
disable_if<has_function<T, f>::value, int>::type call(T const&) {
    std::cout << "has no f" << std::endl;
}

int main() {
    call(has_no_f()); // "has no f"
    call(has_f()); // "has f"
}

仅重载 call 不起作用，因为实际上有很多 foo 和 bar 类型以及 call 函数对此一无所知(基本上 call 位于a内，用户提供自己的类型).

Simply overloading call doesn’t work since there are actually a lot of foo and bar types and the call function has no knowledge of them (basically call is inside a and the users supply their own types).

我无法使用C ++ 0x，并且需要适用于所有现代编译器的解决方案.

I cannot use C++0x, and I need a working solution for all modern compilers.

注意:类似问题的解决方案不幸的是在这里不起作用.

Note: the solution to a similar question unfortunately doesn’t work here.

推荐答案

#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>
#include <functional>
#include <type_traits>

struct X {};
struct Y {};

__int8 f(X x) { return 0; }
__int16 f(...) { return 0; }

template <typename T> typename std::enable_if<sizeof(f(T())) == sizeof(__int8), int>::type call(T const& t) {
    std::cout << "In call with f available";
    f(t);
    return 0;
}

template <typename T> typename std::enable_if<sizeof(f(T())) == sizeof(__int16), int>::type call(T const& t) {
    std::cout << "In call without f available";
    return 0;
}

int main() {
    Y y; X x;
    call(y);
    call(x);
}

快速修改f()的返回类型将产生传统的SFINAE解决方案.

A quick modification of the return types of f() yields the traditional SFINAE solution.

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