通过std ::函数包装重载的函数 [英] Wrap overloaded function via std::function
问题描述
我有一个重载函数,我想传递一个包装在std ::函数。 GCC4.6没有找到匹配函数。
虽然我在这里找到一些问题的答案不是我想要的那样明确。有人能告诉我为什么下面的代码不能扣除正确的重载和如何(优雅地)工作。
int test (const std :: string&){
return 0;
}
int test(const std :: string *){
return 0;
}
int main(){
std :: function< int(const std :: string&)> func = test;
return func();
}
/ p>
要消除歧义,请使用显式强制类型:
typedef int * funtype)(const std :: string&);
std :: function< int(const std :: string&)> func = static_cast< funtype>(test); // cast!
现在编译器能够根据转换的类型来消除这种情况。 p>
或者,您可以这样做:
typedef int(* funtype) (const std :: string&);
funtype fun = test; //现在不需要施放!
std :: function< int(const std :: string&)> func = fun; //没有施法!
所以为什么 std :: function< int(const std :: string& )>
不工作的方式 funtype fun = test
以上工作?
很好的答案是,因为 std :: function
可以用任何对象初始化,因为它的构造函数是模板化的它独立于传递给 std :: function
的模板参数。
I have an overloaded function which I want to pass along wrapped in a std::function. GCC4.6 does not find a "matching function". While I did find some questions here the answers are not as clear as I would like them. Could someone tell me why the following code can not deduct the correct overload and how to (elegantly) work around it?
int test(const std::string&) {
return 0;
}
int test(const std::string*) {
return 0;
}
int main() {
std::function<int(const std::string&)> func = test;
return func();
}
That is ambiguous situation.
To disambiguate it, use explicit cast as:
typedef int (*funtype)(const std::string&);
std::function<int(const std::string&)> func=static_cast<funtype>(test);//cast!
Now the compiler would be able to disambiguate the situation, based on the type in the cast.
Or, you can do this:
typedef int (*funtype)(const std::string&);
funtype fun = test; //no cast required now!
std::function<int(const std::string&)> func = fun; //no cast!
So why std::function<int(const std::string&)>
does not work the way funtype fun = test
works above?
Well the answer is, because std::function
can be initialized with any object, as its constructor is templatized which is independent of the template argument you passed to std::function
.
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