C ++ 11 result_of推导我的函数类型失败 [英] C++11 result_of deducing my function type failed
问题描述
我在尝试下面的程序:
#include<type_traits>
using namespace std;
template <class F, class R = typename result_of<F()>::type>
R call(F& f) { return f(); }
int answer() { return 42; }
int main()
{
call(answer);
return 0;
}
call(answer)无法编译
"call(answer)" fails to compile
VC说'R调用(F&)'不能推导出'R'的模板参数
VC says 'R call(F&)' could not deduce template argument for 'R'
GCC说|注意:模板参数扣除/ replace failed:| error:函数返回一个函数
GCC says |note: template argument deduction/substitution failed:|error: function returning a function
我不知道一个函数名称是否可以用于模板。我在哪里错了,如何使我的电话(答案)工作?
I'm not sure if a "function name" could be used for templates. Where did I get wrong, how to make my call(answer) work?
推荐答案
您可以在这些情况下使用转寄参考:
You can use forwarding references in these cases:
#include<type_traits>
#include<utility>
#include<cassert>
using namespace std;
template <class F, class R = typename result_of<F()>::type>
R call(F&& f) { return std::forward<F>(f)(); }
int answer() { return 42; }
int main()
{
assert(call(answer) == 42);
return 0;
}
通常可以避免麻烦。
也就是说,为什么你的代码不工作很好地解释@TC在其答案中。
更多细节,请参阅此问题的意见。
That said, why your code doesn't work is nicely explained by @T.C. in his answer.
See also the comments to this question for further details.
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