[英]How to define a template function with two relative parameters
I'm trying to define a function, which allows us to call the standard hash function or some custom-defined function, and return the value of hash. I'm trying to define a function, which allows us to call the standard hash function or some custom-defined function, and return the value of hash.
Here is an example about how to use my function:这是一个关于如何使用我的 function 的示例:
auto res = myfunc<std::hash>(2); //hash the integer 2 with the function std::hash
auto res2 = myfunc<std::hash>("abc"); // hash the string "abc" with the function std::hash
auto res3 = myfunc<customHasher>(2); // hash the integer 2 with some custom hash function
I've tried to code as below:我尝试编写如下代码:
template<void (*T)(U)>
size_t myfunc(const U &u)
{
return T<U>(u);
}
T
should be a function pointer, a std::function
or a lambda, and U
is the type of parameter of T
. T
应该是 function 指针、 std::function
或 lambda, U
是T
的参数类型。
But it can't be compiled.但是不能编译。
main.cpp:14:23: error: expected ‘>’ before ‘(’ token
template<void (*T)(U)>
^
main.cpp:15:25: error: ‘U’ does not name a type
size_t myfunc(const U &u)
^
main.cpp: In function ‘size_t myfunc(const int&)’:
main.cpp:17:18: error: ‘U’ was not declared in this scope
return T<U>(u);
Well, I know that template<void (*T)(U)>
must be wrong because U
is not defined.好吧,我知道
template<void (*T)(U)>
一定是错误的,因为U
没有定义。 But I don't know how to fix it.但我不知道如何解决它。
You need to declare both parameters.您需要声明这两个参数。 Moreover,
std::hash
is a class template, not a function.此外,
std::hash
是 class 模板,而不是 function。 You can use a template template parameter:您可以使用模板模板参数:
#include <cstdint>
#include <functional>
#include <iostream>
#include <string>
template<typename T, template<typename> typename H = std::hash>
std::size_t myfunc(const T &t)
{
return H<T>{}(t);
}
int main() {
std::cout << myfunc(std::string{"123"});
}
Though, to use the same with your customHasher
it needs to be a class template (with operator()
), too.不过,要与您的
customHasher
一起使用,它也需要是一个 class 模板(带有operator()
)。
Note that you need to explicitly construct a string in main
, otherwise T
cannot be deduced to be std::string
.请注意,您需要在
main
中显式构造一个字符串,否则T
不能被推断为std::string
。
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