这是此问题的后续解答,发现here!
现在,我可以实例化该对象。我现在收到Visual Studio C2440编译器错误...
在模板化之前的原始代码中,我有一组在std::function<double(double)>
成员对象上工作的成员函数,如下所示:
struct SomeStruct {
double a_;
double b_;
SomeStruct(double a, double b) : a_{a}, b_{b} {}
};
class SomeClass {
private:
SomeStruct fields_;
size_t n_;
std::function<double(double)> func_;
public:
SomeClass(SomeStruct fields, size_t n, std::function<double(double)> func) :
fields_{fields}, n_{n}, func_{func}
{}
double evaluate() {
auto distance = fields_.a_ - fields_.b_;
auto dx = distance / n_;
return calculate(dx, fields_);
}
private:
double calculate(double dx, const SomeStruct& someStruct) {
double result = 0;
for (size_t i = 0; i < n_; ++i) {
auto dy = func_(someStruct.a_ + i * dx);
auto area = dy * dx;
result += area;
}
return result;
}
};
现在我的班级看起来像这样:
template<typename Field>
struct SomeStruct {
Field a_;
Field b_;
constexpr SomeStruct(Field a, Field b) : a_{a}, b_{b} {}
};
template<typename FieldType, typename ValueType, typename Func>
class SomeClass {
private:
SomeStruct<FieldType> fields_;
size_t n_;
Func* func_;
public:
SomeClass(SomeStruct<FieldType> fields, size_t n, Func func) :
fields_{fields}, n_{n}, func_{func}
{}
constexpr ValueType evaluate() {
auto distance = fields_.a_ - fields_.b_;
auto dx = distance / n_;
return calculate(dx, fields_);
}
private:
constexpr ValueType calculate(ValueType dx, const SomeStruct<FieldType>& someStruct) {
ValueType result = 0;
for (size_t i = 0; i < n_; ++i) {
auto dy = func_(someStruct.a_ + i * dx);
auto area = dy * dx;
result += area;
}
return result;
}
};
template<typename FieldType, typename ValueType, typename Func>
SomeClass(SomeStruct<FieldType>, ValueType, Func) ->
SomeClass<FieldType, ValueType, Func>;
我现在正尝试使用此类:
template<typename T>
constexpr T funcA(T x) {
return x;
}
template<typename T>
constexpr T funcB(T x) {
return x*x;
}
int main() {
SomeClass a{SomeStruct{1.0, 3.0}, 1000, &funcA<double>};
// a.evaluate();
SomeClass b{SomeStruct{3.5, 7.5}, 2000, &funcB<double>};
// b.evaluate();
return 0;
}
而且我收到此Visual Studio错误...
1>------ Build started: Project: Computations, Configuration: Debug Win32 ------
1>main.cpp
1>c:\users\...\main.cpp(33): error C2440: 'initializing': cannot convert from 'initializer list' to 'Integrator<double,int,T (__cdecl *)(T)>'
1> with
1> [
1> T=double
1> ]
1>c:\users\...\main.cpp(33): note: No constructor could take the source type, or constructor overload resolution was ambiguous
1>Done building project "Computations.vcxproj" -- FAILED.
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========
这是在我甚至调用其公共evaluate()
函数之前就会生成的,该函数将调用它的私人calculate()
功能...
我正在尝试解决此编译器错误,并且试图确保能够调用正确存储在此类中的函数对象,函数指针,函子,lambda ...
问题出在数据成员Func
的声明中:
Func* func_;
它声明func_
属于pointer,属于推导指南推导的类型:
template <typename FieldType, typename ValueType, typename Func>
SomeClass(SomeStruct<FieldType>, ValueType, Func)
-> SomeClass<FieldType, ValueType, Func>;
通过使用函数的地址调用构造函数:
SomeClass a{SomeStruct{1.0, 3.0}, 1000, &funcA<double>};
Func
推导为:
Func = double(*)(double)
因此,Func*
成为指向函数指针的指针:
Func* = double(**)(double)
并且无法用&funcA<double>
初始化。而是,单独使用Func
作为数据成员类型。
我试图确保能够调用正确存储在此类中的函数对象,函数指针,函子,lambda
要么为此使用概念,要么使用静态断言:
static_assert(std::is_invocable_v<Func&, FieldType&>, "Invalid callable");