[英]How to write a template wrapper method for other class member functions?
我嘗試使用不同的參數為不同的函數創建模板化包裝器。 設置是A
類,基本實現了兩個方法foo
和bar
。 另一個B
類應包裝這些方法並添加新功能。
以下鏈接的解決方案非常適用於非類函數: c ++ 11:模板化包裝函數
但是如果我嘗試從另一個類調用方法,我會收到錯誤。
#include <algorithm>
#include <functional>
#include <iostream>
class A
{
public:
void foo(int x) {
std::cout << "Foo: " << x << std::endl;
}
void bar(int x, float y) {
std::cout << "Bar: " << x << ", " << y << std::endl;
}
};
class B
{
public:
void fooAndMore(int x) {
foobarWrapper(&A::foo, 1);
}
void barAndMore(int x, float y) {
foobarWrapper(&A::bar, 1, 3.5f);
}
template<typename T, typename... Args>
void foobarWrapper(T&& func, Args&&... args)
{
std::cout << "Start!" << std::endl;
std::forward<T>(func)(std::forward<Args>(args)...);
std::cout << "End!" << std::endl;
}
};
int main()
{
B b;
b.fooAndMore(1);
b.barAndMore(2, 3.5f);
}
我期待這樣的事情:
Start!
Foo: 1
End!
Start!
Bar: 1, 3.5
End!
但相反,我得到:
error C2064: term does not evaluate to a function taking 1 arguments
note: see reference to function template instantiation 'void B::foobarWrapper<void(__thiscall A::* )(int),int>(T &&,int &&)' being compiled
with
[
T=void (__thiscall A::* )(int)
]
error C2064: term does not evaluate to a function taking 2 arguments
note: see reference to function template instantiation 'void B::foobarWrapper<void(__thiscall A::* )(int,float),int,float>(T &&,int &&,float &&)' being compiled
with
[
T=void (__thiscall A::* )(int,float)
]
不知道怎么解決這個問題?
最簡單的解決方法是使A
類的成員函數是static
。 ( 見在線 )
class A
{
public:
static void foo(int x) {
^^^^^^
std::cout << "Foo: " << x << std::endl;
}
static void bar(int x, float y) {
^^^^^^
std::cout << "Bar: " << x << ", " << y << std::endl;
}
};
否則,您需要傳遞A
類的實例以在foobarWrapper
函數中調用其成員函數。 使用lambdas,您可以將它們打包到可調用的func
並傳遞給foobarWrapper
。
( 見在線 )
class B
{
public:
void fooAndMore(const A& a_obj, int x) {
foobarWrapper([&]() { return a_obj.foo(x); });
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^Args captured to the lambda
}
void barAndMore(const A& a_obj, int x, float y) {
foobarWrapper([&]() { return a_obj.bar(x, y); });
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Args captured to the lambda
}
template<typename T>
void foobarWrapper(T&& func) // no Args needed any more (@credits Jarod42)
{
std::cout << "Start!" << std::endl;
std::forward<T>(func)(); // simply call the func
std::cout << "End!" << std::endl;
}
};
int main()
{
B b;
b.fooAndMore(A{}, 1); // pass a temporary A object
b.barAndMore(A{}, 2, 3.5f);
}
試試這個,
#include <algorithm>
#include <functional>
#include <iostream>
class A
{
public:
void foo(int x) {
std::cout << "Foo: " << x << std::endl;
}
void bar(int x, float y) {
std::cout << "Bar: " << x << ", " << y << std::endl;
}
};
class B
{
public:
void fooAndMore(int x) {
foobarWrapper(&A::foo, x);
}
void barAndMore(int x, float y) {
foobarWrapper(&A::bar, x, y);
}
template<typename T, typename... Args>
void foobarWrapper(T func, Args&&... args)
{
std::cout << "Start!" << std::endl;
auto caller = std::mem_fn( func); // Newly added lines
caller( A(), args...); // Newly added line
std::cout << "End!" << std::endl;
}
};
int main()
{
B b;
b.fooAndMore(1);
b.barAndMore(2, 3.5f);
}
輸出:
Start!
Foo: 1
End!
Start!
Bar: 2, 3.5
End!
有關詳細信息,請參閱此鏈接std :: mem_fn
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