Overloaded pointer to function
c++
Solution
1) It's not necessary, really. If you have used `using namespace std` directive, it's necessary to cast to the desired type to let the compiler know which overload you want. So, you might also say
transform(begin(s), end(s), back_inserter(result), static_cast<int(*)(int)>(&toupper));
Otherwise the following should be enough:
transform(begin(s), end(s), back_inserter(result), ::toupper);
2) Identifiers that are function names decay into pointers, yes, but they aren't exactly the same thing. That being said, in this case it should be fine to say
int (*toupperp)(int) = toupper;
or even (if you haven't used `using namespace std` directive):
auto toupperp = toupper;
3) it's for compatibility with C standard library. It's used on every element of `s`, which for `string` is a `char`.
Problem
I was looking over the following code: ``` string toUpper(string s) { string result; int (*toupperp)(int) = &toupper; // toupper is overloaded transform(begin(s), end(s), back_inserter(result), toupperp); return result; } ``` I am confused by this line: ``` int (*toupperp)(int) = &toupper; // toupper is overloaded ``` 1.Why is this line necessary? 2.I believe that `&` retrieves a pointer to something from memory. But `toupper`, the name of the function is already a pointer, no? Why can't we do this: ``` int (*toupperp)(int) = toupper; ``` 3.Why is the function overloaded to `int` if it's used on a `string`?