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C++：对象判断是否含有某个成员函数

 3 years ago
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C++：对象判断是否含有某个成员函数 - Ryan_naiquan

C++

sizeof在编译期执行，结果是函数返回类型的大小，函数并不会被调用。sizeof获取返回值的函数可以不定义。

#include <iostream>

template<typename T>
struct has_no_destroy {
	template<typename C>
	static char test(decltype(&C::no_destroy));


	template<typename C>
	static int32_t test(...);

	const static bool value = sizeof(test<T>(0)) == 1;
};
// 其作用就是用来判断是否有 no_destroy 函数

struct A {

};

struct B {
	void no_destroy() {}
};
struct C {
	int no_destroy;
};

struct D : B {

};

int main()
{
	printf("%d\n", has_no_destroy<A>::value);
	printf("%d\n", has_no_destroy<B>::value);
	printf("%d\n", has_no_destroy<C>::value);
	printf("%d\n", has_no_destroy<D>::value);
	system("PAUSE");
	return 0;
}

本文由 Ryan 创作，采用 知识共享署名4.0 国际许可协议进行许可
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最后编辑时间为: 2021/05/18 15:54


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