[英]mem_fun + bind2nd allows to call a method with an argument of an arbitrary type
考虑这个例子( https://ideone.com/RpFRTZ )
这将使用不相关类型Bar
的参数有效地调用Foo::comp (const Foo& a)
。 如果我注释掉std::cout << "a = " << as << std::endl;
它也以某种方式工作和打印Result: 0
如果我打印出值,而不是段错误,这是公平的......但为什么它首先编译?
#include <functional>
#include <string>
#include <iostream>
struct Foo
{
bool comp(const Foo& a)
{
std::cout << "a = " << a.s << std::endl;
return a.s == s;
}
std::string s;
};
struct Bar
{
int a;
};
template <class F, class T>
void execute (F f, T a)
{
std::cout << "Result: " << f (a) << std::endl;
}
int main()
{
Foo* f1 = new Foo;
f1->s = "Hello";
Foo f2;
f2.s = "Bla";
Bar b;
b.a = 100;
execute (std::bind2nd (std::mem_fun(&Foo::comp), b), f1);
return 0;
}
答案是在std :: bind2nd的实现中:
template<typename _Operation, typename _Tp>
inline binder2nd<_Operation>
bind2nd(const _Operation& __fn, const _Tp& __x)
{
typedef typename _Operation::second_argument_type _Arg2_type;
return binder2nd<_Operation>(__fn, _Arg2_type(__x));
}
您可以看到右侧类型存在不安全的C样式转换“_Arg2_type(__ x)”,因此您的示例编译就像编写它一样:
execute (std::bind2nd (std::mem_fun(&Foo::comp), (const Foo&)b), f1);
这是不幸的有效C ++代码。
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