CODEWITHSUNDEEP
c++functionclassarguments
CChiste
CChiste

Reputation: 131

Passing a function as a parameter CPP

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.

Upvotes: 3

Views: 1063

Answers (4)

Zac Howland
Zac Howland

Reputation: 15872

There is no need for you to reinvent the wheel. ::toupper and ::tolower take and return int, but their valid range is that of unsigned char. Additionally, std::toupper and std::tolower both take char.

Since it appears you are not using std::string, I'll try to keep it as close to your code as possible:

void upperList(LineEditor line)
{
    char c;
    // you do not have a begin() function??
    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();
        c = std::toupper(c);
        line.at() = c; // assuming this returns a reference
        line.right(); 
    }
}

It becomes much easier if you modify your string class to behave more like the std::string class:

std::string line;
std::transform(line.begin(), line.end(), line.begin(), std::ptr_fun<int, int>(std::toupper));

Example

Upvotes: 0

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

the type of expression ToUpper(c) is char. So call

line.apply(ToUpper(c));

means to call function apply with argument of type char.

You should define the function as

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

Upvotes: 0

Roland Illig
Roland Illig

Reputation: 41625

The expression ToUpper(c) calls the function, but when calling apply you don't want to call that function immediately, so you need do say apply(ToUpper), since ToUpper is the way to access the function itself.

Upvotes: 0

David G
David G

Reputation: 96810

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).

Upvotes: 2

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