Thread: function as parmeter

Code:

double f(double a){return a;} 
int main(void) {
  double d1,d2; 
derivatives(f(1.0),1, 0.0001,&d1,&d2); 
  
    return (0); 
} /* end main */

Code:

int derivatives(double f(double a), double x, double eps, double * d1, double * d2){
    
    (*d1)=(f(x+eps)-f(x-eps))/(2*eps);
    printf("%lf",*d1);
    (*d2)=(f(x+eps)-2*f(x)+f(x-eps))/(eps*eps);

    
}

error says incomtabile for parameter 1
any help please ?

In the standard C language, you can not insert functions into parameters, use macros and defines.

Originally Posted by Washington

In the standard C language, you can not insert functions into parameters, use macros and defines.

C supports passing a function pointer to another function, which seems to be the intent here.

Change "double f(double a)" to "double (*f)(double a)", and then change "f(1.0)" in the call to derivatives to just "f".

Your code should look more like this:

Code:

int derivatives(double (*f)(double a), double x, double eps, double * d1, double * d2);

derivatives(f,1, 0.0001,&d1,&d2);

You can use a pointer to a function and also pass the parameter(s) which go with that function.
It does not look nice/readable but it can be done.
For example:

Code:

#include <stdio.h>

void doWhat(int (*fPtr) (), int param1)
{
    printf("%d\n", (*fPtr)(param1));
}

int intFunc(int a)
{
    return a * 2;
}

int main(void)
{
    int (*pt_iFunc) ();

    pt_iFunc = intFunc;

    doWhat(pt_iFunc, 5);

    return (0);
}

sorry christop, while I was typing you already sent an answer

You don't actually *need* the weird declaration for the function pointer. The following is 100% valid C:

Code:

#include <stdio.h>

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

int bar(int func(int), int x)
{
    return func(x);
}

int main(void)
{
    printf("%d\n", bar(foo, 2));
    return 0;
}

Code:

> gcc --std=c89 -Wall -pedantic blah.c 
> 
> ./a.out
4
>

Without looking too much into it I think the OP's problem was having

Code:

int derivatives(double f(double a), double x, double eps, double * d1, double * d2)

Rather than

Code:

int derivatives(double f(double), double x, double eps, double * d1, double * d2)

(not that in the first instance it names the parameter and in the second one that should work it only has the type

Originally Posted by Hodor

You don't actually *need* the weird declaration for the function pointer.

Wow, you're right! I honestly didn't know that, though I personally prefer using the "weird" declaration because it's the same syntax you have to use when you declare a function pointer in any other context.

Originally Posted by Hodor

Without looking too much into it I think the OP's problem was having

Code:

int derivatives(double f(double a), double x, double eps, double * d1, double * d2)

Rather than

Code:

int derivatives(double f(double), double x, double eps, double * d1, double * d2)

(not that in the first instance it names the parameter and in the second one that should work it only has the type

Giving the function pointer's parameter a name isn't an error, as I just tested it with "gcc -Wall -std=c89 -pedantic". The only problem in the OP's code was in the invocation of the "derivatives" function.