Passing a function as a parameter CPP

Question

I am attempting to call a class function in my main program that takes a function as its parameter, and applies the function to a private list. I am getting the error invalid conversion from char to char (*f)(char). Hopefully I just don't understand how to pass functions as paremeters. The following are functions in my main cpp file

char ToUpper(char c)
{
char b='A';
for(char a='a';a<='z';a++)
{
   if(a==c)
  {
     c=b;
     break;
  }
  ++b;
}
return c;
}

void upperList(LineEditor line)
{
char c;
for(int i=0;i<100;i++)   //ensure iterator is at beginning of line
  line.left();           

for(int i=0;i<100;i++)
{
  c=line.at();               //assign character current element pointed to by iterator
  line.apply(ToUpper(c));    //problem: trying to apply ToUpper function to char c
  line.right();              //apply function and increment iterator
}
}

And this is the apply member function

void LineEditor::apply(char (*f)(char c))
{
*it=f(c);
}

Also, in case it wasn't obvious, I tried using the cctypes toupper and tolower but they take and return integers.

Answer 1

When you call ToUpper, it doesn't return the function, it returns the (supposed) character in its uppercase form.

Another reason this doesn't work is because you cannot create arguments inside the signature of a function pointer. The area for the parameter only designates the type that the function takes. This...

char (*f)(char c);
//        ^^^^^^

is therefore wrong.

Solution:

Use a std::function and std::bind it to an argument:

#include <functional>

line.apply(std::bind(ToUpper, c));

It requires the signature of apply to be changed to:

void LineEditor::apply(std::function<char (char)> f);

If you can't do this, you can simply let apply take a second parameter as the argument:

void LineEditor::apply(char (*f)(char), char c);

and call it as apply(ToUpper, c).