function as a input parameter

**GaPe** · 02-02-2003

Hi!

Can someone please help me with this code? I'm lost. I get one warning and two error messages.

".../problem.c(19): warning C4047: 'function' : 'int (__cdecl *)(int,int)' differs in levels of indirection from 'int'"

".../problem.c(29): error C2065: 'a' : undeclared identifier"

".../problem.c(29): error C2065: 'b' : undeclared identifier"

Code:

# include <stdio.h>
# include <conio.h>


int  Multiply (int a, int b);
void WriteOut (int (*Multiply) (int a, int b));
void WaitForKey (void);


int main ()
{
	int a = 0, b = 0;


	printf ("Enter first number: ");
	scanf ("%d", &a);
	printf ("Enter second number: ");
	scanf ("%d", &b);
	WriteOut ( Multiply (a, b) );
	WaitForKey ();
	return 0;  
}
int Multiply (int a, int b)
{
    return (a * b);
}
void WriteOut (int (*Multiply) (int a, int b))
{
    printf ("Result = %d", Multiply (a, b));
}

**Shiro** · 02-02-2003

Your function WriteOut expects a pointer to function only, you cannot pass a and b the way you currently do. If you want the variables also be passed, pass them seperately.

Prototype WriteOut as:
void WriteOut (int (*Multiply) (int, int), int a, int b);

And use it as:
WriteOut (Multiply, a, b);

**MisterBadger** · 02-02-2003

Hi GaPe

I used your Multiply function with printf to display a*b.

Do you really need the Multiply function? This line would give the same result:

printf("\nProduct of %d and %d is %d\n\n",a,b, a*b);

Here's my code:

Code:

# include <stdio.h>
# include <conio.h>

int  Multiply (int a, int b);
//void WriteOut (int (*Multiply) (int a, int b));
//void WaitForKey (void);

int main ()
{
	int a = 0, b = 0;


	printf ("Enter first number: ");
	scanf ("%d", &a);
	printf ("Enter second number: ");
	scanf ("%d", &b);
//	WriteOut ( Multiply (a, b) ); // I used this code to display the result:

printf("\nProduct of %d and %d is %d\n\n",a,b, Multiply(a,b));

//	WaitForKey (); // This function was not defined.
		       // Does the following code do what you intended...
                       // Program will wait for next keystroke...
   printf("Any key to clear screen and end prog > ");
	getch();
	clrscr();
	printf("END OF PROG!\n\n");
	return 0;
}
int Multiply (int a, int b)
{
	 return (a * b);
}

/*void WriteOut (int (*Multiply) (int a, int b))
{
	 printf ("Result = %d", Multiply (a, b));
} I didn't use this function...*/

Cheers from Badger :-)

**Shiro** · 02-02-2003

Gape probably wants to do some generic programming. With use of a function pointer, he can also pass other functions of the same type to WriteOut, like

int sum (int a, int b);
int subtract (int a, int b);
int divide (int a, int b);
etc.

**GaPe** · 02-02-2003

MisterBadger you're totally off of what I want.

Shiro is right. I just want to know how to use function as a input parameter in other function. So please help me out with this.

One more thing. What are the advantages/disadvantages for that kind of use of functions?

**Hammer** · 02-02-2003

>>I just want to know how to use function as a input parameter in other function. So please help me out with this.
Shiro already gave you the answer. Here's a working version of your code:

Code:

# include <stdio.h>
# include <conio.h>


int  Multiply (int a, int b);
void WriteOut (int (*Multiply) (int, int), int, int);
void WaitForKey (void);


int main ()
{
	int a = 0, b = 0;


	printf ("Enter first number: ");
	scanf ("%d", &a);
	printf ("Enter second number: ");
	scanf ("%d", &b);
	WriteOut ( Multiply, a, b );
	//WaitForKey ();
	return 0;  
}
int Multiply (int a, int b)
{
    return (a * b);
}
void WriteOut (int (*Multiply) (int, int), int a, int b)
{
    printf ("Result = %d", Multiply (a, b));
}

Thread: function as a input parameter

Thread Tools

Search Thread

Display

function as a input parameter

Similar Threads

Getting an error with OpenGL: collect2: ld returned 1 exit status

Consequence of const-ifying a function parameter

<Gulp>

Problem with Visual C++ Object-Oriented Programming Book.

Custom Made Safe Input Function