conditional function

[bertazoid]

Is there anyway to define a function conditionally? For example in psuedo code:

IF (condition 1) then function(x)=x^2 ELSE function (x) = x.

I tried defining the function within an if staement, and it didn't work. I also thought about using the if statement within the definition for the function, but then is it possible to use variables/call other functions from 'main'? i.e. in this case I would need to pass 'condition 1' from main to the if statement within 'function'...(or condition 1 uses variables from main).

[tabstop]

The most likely option, it seems to me, would be to define two functions, calling the appropriate one based on your if. Is there some reason why this isn't an option? (Edit: Or, I suppose, if the function call is going to happen later/repeatedly, then you could use a function pointer to set up the appropriate function to use (again, this would still require defining two functions).)

[MeTh0Dz]

What you could do is just use one function, and as one of your parameters pass an flag that will allow the function to run one if statement or the other.

Example.

void Ex_Funt(int var1, char * var2, int flag) {
     if (flag == 0) {
          /* ... */
     }
     else if (flag == 1) {
          /* ... */
     }
     else {
          /* ... */
     }
}

Another option could be using two functions and then just a function pointer, or just call the function.

Just a couple possible solutions.

[Dino]

You can use preprocessor directives to generate your function however(s) however you wish. #if is what you want to use.

[bertazoid]

Well I think I could use two functions to do what I want, and then just use each one as appropriate in the loop. I had a look at function pointers, but didn't really understand it...will have to save that for another day.

Thanks

[Dave_Sinkula]

Function pointers may or may not be overkill.

#include <stdio.h>

int foo(int x)
{
   return x * x;
}

int bar(int x)
{
   return x;
}

int test(int x, int i)
{
   int (*function)(int) = i ? foo : bar;
   return function(x);
}

int main(void)
{
   int i, a = 5;
   for (i = 0; i < 2; ++i)
   {
      printf("test(&#37;d,%d) = %d\n", a, i, test(a,i));
   }
   return 0;
} 

/* my output
test(5,0) = 5
test(5,1) = 25
*/

[Dino]

Here's a function pointer example - it's really not that difficult a concept:

#include <stdio.h>

int foo(int x) {
	return x * x;
}

int bar(int x) {
	return x;
}

int main(void) {

	int ( *myfunc ) () ; 
   
	myfunc = foo ; 
	printf("myfunc(5) = &#37;d\n", myfunc(5) );

	myfunc = bar ; 
	printf("myfunc(5) = %d\n", myfunc(5) );
   
	return 0;
}

Output:

myfunc(5) = 25
myfunc(5) = 5

[bertazoid]

Thanks for the examples I'm understading it now, and have my program working (this part at least).