[英]Template type deduction with std::function
I have discovered the following behaviour with std::function
and type deduction, which was unexpected for me: 我用
std::function
和type deduction发现了以下行为,这对我来说意外:
#include <functional>
template <typename T>
void stdfunc_test(std::function<T(T)> func) {};
int test_func(int arg)
{
return arg + 2;
}
int main()
{
stdfunc_test([](int _) {return _ + 2;});
stdfunc_test(test_func);
}
Both lines in main
result in error: main
中的两行都会导致错误:
no instance of function template "stdfunc_test" matches the argument list
没有函数模板“stdfunc_test”的实例与参数列表匹配
When attempting to compile in Visual Studio 2015. 尝试在Visual Studio 2015中编译时。
Why doesn't the type deduction deduct template type from the function type, and is there a workaround for it? 为什么类型扣除不会从函数类型中扣除模板类型,是否有解决方法?
You can use templates to deduce the signature of functions and functors: 您可以使用模板来推断函数和仿函数的签名:
#include<functional>
template<class T>
struct AsFunction
: public AsFunction<decltype(&T::operator())>
{};
template<class ReturnType, class... Args>
struct AsFunction<ReturnType(Args...)> {
using type = std::function<ReturnType(Args...)>;
};
template<class ReturnType, class... Args>
struct AsFunction<ReturnType(*)(Args...)> {
using type = std::function<ReturnType(Args...)>;
};
template<class Class, class ReturnType, class... Args>
struct AsFunction<ReturnType(Class::*)(Args...) const> {
using type = std::function<ReturnType(Args...)>;
};
template<class F>
auto toFunction( F f ) -> typename AsFunction<F>::type {
return {f};
}
template <typename T>
void stdfunc_test(std::function<T(T)> func) {};
int test_func(int arg)
{
return arg + 2;
}
int main()
{
stdfunc_test( toFunction([](int _) {return _ + 2;}) );
stdfunc_test( toFunction(test_func) );
return 0;
}
you can try it live here: http://fiddle.jyt.io/github/d4ab355eb2ab7fc4cc0a48da261f0127 你可以在这里试试吧: http : //fiddle.jyt.io/github/d4ab355eb2ab7fc4cc0a48da261f0127
No implicit conversion is performed during template argument deduction, except: temp.deduct.call 在模板参数推导期间不执行隐式转换,除了: temp.deduct.call
In general, the deduction process attempts to find template argument values that will make the deduced A identical to A (after the type A is transformed as described above).
通常,演绎过程试图找到模板参数值,这些参数值将使推导出的A与A相同(在如上所述变换类型A之后)。 However, there are three cases that allow a difference:
但是,有三种情况可以产生差异:
- If the original P is a reference type, the deduced A (ie, the type referred to by the reference) can be more cv-qualified than the transformed A.
如果原始P是引用类型,则推导出的A(即,引用所引用的类型)可以比转换后的A更具有cv限定。
- The transformed A can be another pointer or pointer to member type that can be converted to the deduced A via a function pointer conversion ([conv.fctptr]) and/or qualification conversion ([conv.qual]).
变换后的A可以是另一个指向成员类型的指针或指针,可以通过函数指针转换([conv.fctptr])和/或限定转换([conv.qual])将其转换为推导出的A.
- If P is a class and P has the form simple-template-id, then the transformed A can be a derived class of the deduced A. Likewise, if P is a pointer to a class of the form simple-template-id, the transformed A can be a pointer to a derived class pointed to by the deduced A.
如果P是一个类而P的形式为simple-template-id,则转换后的A可以是推导出的A的派生类。同样,如果P是指向simple-template-id形式的类的指针,变换A可以是指向由推导出的A指向的派生类的指针。
However, if the template parameter doesn't participate in template argument deduction, implicit conversion will be performed: ( temp.arg.explicit ) 但是,如果模板参数不参与模板参数推导,则将执行隐式转换:( temp.arg.explicit )
Implicit conversions (Clause [conv]) will be performed on a function argument to convert it to the type of the corresponding function parameter if the parameter type contains no template-parameters that participate in template argument deduction.
如果参数类型不包含参与模板参数推导的模板参数,则将对函数参数执行隐式转换(Clause [conv])以将其转换为相应函数参数的类型。 [ Note: Template parameters do not participate in template argument deduction if they are explicitly specified.
[注意:如果明确指定模板参数,则模板参数不参与模板参数推导。
So, if you explicitly specify the template argument, it should work: 因此,如果您明确指定模板参数,它应该工作:
stdfunc_test<int>([](int _) {return _ + 2;});
stdfunc_test<int>(test_func);
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