Let's say I have the following struct to store a reference to a class member variable:
template <typename R>
struct Foo {
R ref;
int info;
};
class Bar {
void* run();
}
I want to create a const variable of type foo that immediately sets the ref param as follows:
const Foo theVar {&Bar::run, 0x1000}; //Doesn't work, but preferred
const Foo<void (Bar::*)()> theVar {&Bar::run, 0x1000}; //Does work, but rather not
The reason I'm using a template in this case is because there's no way to cast a class member variable to a void*, so I'm forced into this position. However, it seems that I cannot declare the variable without first telling the compiler what type I'm planning to use. Although I do have this information, it's not the prettiest way to accomplish what I want and could possibly cause some issues in the long run.
The compiler should be able to tell that I'm passing a variable of type void (Bar::*)() to this struct, so I'm almost certain there has to be a way around this issue.
The reason I'm in need of this struct is because I need a reference to exist for the linker. I don't want it to be set on run-time, it needs to be available for the linker. Using templates seems to be the only way to accomplish this.
CodePudding user response：
The solution is simple: add a deduction guide after Foo declaration:
template<typename R>
struct Foo {
R ref;
int info;
};
template<typename R> Foo(R, int) -> Foo<R>;