将函数作为参数CPP传递

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) .

Answer 2

表达式ToUpper(c)调用该函数，但是当调用apply你不想立即调用该函数，所以你需要说apply(ToUpper) ，因为ToUpper是访问函数本身的方法。

Answer 3

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);
}

Answer 4

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