为什么不是std :: result_of< int（int）> :: type valid？ [英] Why isn't std::result_of<int(int)>::type valid?
问题描述
我已阅读以下相关问题：
I have read the following related questions:
- std::result_of simple function
- decltype, result_of, or typeof?
和 cppreference.com上的 std :: result_of 页面。
所有这些都似乎表明我应该能够使用：
All of them seem to indicate that I should be able to use:
std::result_of<int(int)>::type v1 = 10;
However, when I tried building the following program using g++ 4.9.2我收到LINE A和LINE B的错误消息。错误消息是：
#include <type_traits>
int foo()
{
return 0;
}
int main()
{
std::result_of<int(int)>::type v1 = 10; // LINE A
std::result_of<decltype(foo)>::type v2 = 20; // LINE B
return 0;
}
I get error messages for "LINE A" and "LINE B". The error messages are:我用于编译的命令：
The command I used to compile:FWIW，使用
FWIW, using没有影响。
似乎我不明白应该如何使用 result_of 。
didn't make a difference.
你能解释为什么我
推荐答案
正如你已经发布的链接中所述， result_of 必须是可调用类型或对函数的引用。
As stated in the link you have already posted, the first part of the argument to result_of must be a callable type or a reference to a function.
假设您有
struct Callable
{
int operator()(double);
void operator()(int);
};
然后 result_of 可帮助您确定返回类型，如果你知道参数的类型。对于上述示例：
then result_of helps you determining the return type, if you know the type of the arguments. For the example above:
result_of<Callable(int)>::type == void ... per definition
result_of<Callable(double)>::type == int ... per definition
result_of<Callable(char)>::type == void ... the int-overload matches better
result_of<Callable(float)>::type == int ... the double-overload matches better
$ b b
为了找到函数 foo 的返回类型，你必须通过函数引用：
In order to find the return type of the function foo you would have to go via a function reference:
result_of<decltype(foo)& ()>::type == int
但这似乎有点扭曲，因为你可以直接写
But this seems a bit twisted as you could directly write
decltype(foo()) == int
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