How do I run a function within a function arguement? C/C++

RHLinuxGUY · 05-12-2006, 03:13 PM

I have been wanting to implement a function within a function arguement lately for a couple reasons. Example would be:

void Function1(int 1, int 2, Function2){ //do something }

But as far as I know you have to declare the variable type in the arguement before you can pass anything. If it makes it easier, Function2 is just a getline function. i.e. getline(myfile,someline);
I'm trying to get information from a text file into a part of my program/game that displays fonts onto a graphical window. Instead of ints and chars, what would I put in place of Function2 when void Function is made? What should I do?

--EDIT--
I have seen programs use the return function. Such as this found at cplusplus.com's tutorial...

// function example
#include <iostream>
using namespace std;

int addition (int a, int b)
{
int r;
r=a+b;
return (r);
}

int main ()
{
int z;
z = addition (5,3);
cout << "The result is " << z;
return 0;
}

is that what I would have to do, something similar to this? Though it still does not answer my question as to what to initialize my function arguement as.

spooon · 05-12-2006, 03:23 PM

Google for "function pointers". You can pass a pointer of Function2 as an argument to Function1; but you need to know the types and stuff of Function2 beforehand, and the actual syntax for function pointers is a little tricky.

I am not really sure what you are trying to do though; so can you explain what you want to do more clearly; because function pointers are not usually used that often.

graemef · 05-12-2006, 04:11 PM

You can certainly pass a function (function pointer) to another function but as spoon suggests they are tricky and you *probably* don't need it. Because they are difficult to implement correctly one really good resource which I use is the C++FAQ

ioerror · 05-12-2006, 05:17 PM

Ah, good old function pointers, what could be simpler? Use them all the time. The easiest way to handle the syntax is to use a typedef.

exvor · 05-12-2006, 09:26 PM

From what i can tell from the post im not sure function pointers is what hes after.

I think what you mean is use a function that returns a value and use its return as a data value of another function.

like this

Code:

#include<stdio.h> 

int getnum(void);
void printnum(int); 

int main() 
{
   printnum(getnum()); 
 return 0; 
 
}
int getnum(void)
{
   int numb; 

   numb = 4; 

   return numb; 
}
void printnum(int number) 
{
  printf("%d",number);
}

mabye ?

RHLinuxGUY · 05-12-2006, 11:52 PM

Thank you exvor, that is what I was looking for... though the example code you supplied didn't work for me, I got the basic idea and made something very similar to that:

Code:

#include <iostream>

using namespace std;

int Func1(void)
{
	int a;
	a = 3;
	return a;
}

void Func2(int Value)
{
	cout<<Value<<endl;
}

string Func3(void)
{
	string a = "hello";
	return a;
}

void Func4(string Value)
{
	cout<<Value<<endl;
}

int main() 
{
 
	if(Func1()==2)
	{
		cout<<"It is correct!\n";
	}
	else
	{
		cout<<"It is not!\n";
	}

	Func2(Func1());

	Func4(Func3());
 
 return 0; 
 
}

Strangely, if I make Func3 return a char (thats saying I make it char Func3(void)) it cannot return chars... why?

exvor · 05-13-2006, 12:34 AM

In your example function 3 is supposed to return a string. Actually it gets real complicated if you attempt to return a string this way i belive you need like a pointer to a pointer and such. I generally avoid them. Good news is you dont have to do it this way you can just make the string outside of the function and pass it by refrence to the function and after the function is done modifying the information it will stay there.

only other thing is you would need to make it an array instead of a constant string.

Ohh reason my code above dident work is cause it is C and needs to be compiled with gcc also i was missing a , in there as well. I fixed it i hope.

RHLinuxGUY · 05-13-2006, 01:04 AM

Yes, I cought the comman (,) but it still would spit garbage out after I compiled it.. successfully. :> Thank you again though, I'll look into the string problem you brought up to me. Would I get the same problem with chars? Or should I just stick to what you said about have an external (I'm assuming global, or a pointer) string initialized/defined somehwere else.

ioerror · 05-13-2006, 04:39 AM

Ah, nested function calls. Yeah, you don't need function pointers for that, should have read the post more closely.

Your string problem is that you're returning a local variable which no longer exists when the function returns. You're probably better off defining the string in main and passing it as an argument to Func3.

Also, as exvor pointed out, you don't really want to return a 'string' in Func3, better to return a pointer or reference. Well, in C you'd use a pointer, with C++, a reference is probably easier.

So, something like this:

Code:

string& Func3(string& a)
{       
        a = "hello";
        return a;

}

int main()
{       
        string a;
        ...

        Func4(Func3(a));
}

schneidz · 05-16-2006, 02:16 PM

um, hello

Code:

printf("hello my name is %s", getname());

printf is a function defined in stdio.h

ioerror · 05-16-2006, 03:44 PM

This is an example about nested function calls, as in foo(bar()), hence the trivial code.

P.S. And it's C++, not C.

schneidz · 05-18-2006, 10:55 AM

i apologize for being obnoxious,

thanks for the correction io.

______________

to further clarify in words:

a function can be used as transparently as a variable as long as it returns something and that something is in the format that the other function is expecting.

it's a shorthand way of declaring a new local variable, storing the return of a function to that local variable then passing that variable to another function.

ta0kira · 05-19-2006, 02:29 PM

I use templates; that way you don't have to get the pointer type exactly, and it also works for functions with default arguments:

Code:

#include <iostream>
#include <string>

void ShowMe1(const std::string &sShow)
{ std::cout << sShow << "\n"; }

int ShowMe2(const char *sShow1, const char *sShow2 = "\n")
{
        std::cout << sShow1 << sShow2;
        return 0;
}


template <class Function>
void ShowAnything(const char *sShow, Function *fFunction)
{ (*fFunction)(sShow); }


int main()
{
        ShowAnything("Hello, this is ShowMe1", &ShowMe1);
        ShowAnything("Hello, this is ShowMe2", &ShowMe2);
        return 0;
}

ta0kira

PS This will also work for functors; classes which have an 'operator ()'.