Array of Array of function pointers in C

Question

Presently i am using a function pointer array. But i need to extend it as an array of array of function pointers.The pseudo code is as below:

#include<stdio.h>


int  AnalogValue1( int a)
{
    printf("\nF: AV1 %d", a);
    return 1;
}


int  AnalogValue2( int a)
{
    printf("\nF: AV2 %d", a);
    return 1;
}

int  BinaryValue1( int a)
{
    printf("\nF: BV1 %d", a);
    return 1;
}


int  BinaryValue2( int a)
{
  printf("\nF: BV2 %d", a);
  return 1;
}
int ( *Read_AV_Props[] )(int a)
=
{
 &AnalogValue1,
 &AnalogValue2,
};

int ( *Read_BV_Props[] )(int a) =
{
 &BinaryValue1,
 &BinaryValue2
};

void main()
{
    Read_AV_Props[0](55);   
    //*(Read_Ob_Props[0])(20);

}

Here need one array of array of function pointers so that it can take the address of Read_BV_Props and Read_AV_Props. And from the main code i can call that array of array of function pointers. Is this possible in C?

Filipe Gonçalves · Accepted Answer · 2014-01-09 13:10:11Z

1

This is the declaration you want:

int (**myArray[])(int) = {
    Read_AV_Props,
    Read_BV_Props
};

This is an array of pointers to pointers to functions receiving int and returning int. It's not a true 2D array because Read_AV_Props and Read_BV_Props decay into a pointer to their first element, i.e., a pointer to pointer to function receiving int and returning int.

You can use it like this:

int a = (**(myArray[i]+j))(1);

Or, due to array-to-pointer decayment rule (in expressions), you could also use:

int a = (*myArray[i][j])(1);

If you want a true 2D array, you must initialize it all in one place:

int (*myArray[][])(int) = {
    { &binaryValue1, &binaryValue2 },
    { ... }
};

The usage is similar.

Or you can make an array of pointers to array of pointer to function receiving int and returning int with:

int (*(*myArray[])[sizeof(Read_AV_Props)/sizeof(*Read_AV_Props)])(int) = {
    &Read_AV_Props,
    &Read_BV_Props
};

But this assumes that Read_AV_Props and Read_BV_Props have the same size.

You can use this with:

int a = (*(*myArray[i])[j])(1);

Given your question, I believe the first option is what you're looking for.

edited Jan 9, 2014 at 13:10

answered Jan 9, 2014 at 12:46

Filipe Gonçalves

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2 Comments

Ginu Jacob Over a year ago

Thanks, your 1st set of code is the exact match but as you said it is getting decayed to just the first pointer of the list...

Filipe Gonçalves Over a year ago

@GinuJacob You're welcome. Nice question btw, I love this pointer stuff ;)

Devolus · Accepted Answer · 2014-01-09 13:09:47Z

0

If you need a 2D array, then you can do it like this:

#include <stdio.h>
#include <stdlib.h>

int func1(int a)
{
    return a*2;
}

int func2(int a)
{
    return a*3;
}

typedef int (*FunctionPtrType)(int a);

FunctionPtrType myArray[2][2] =
{
    // func1, func2
 { func1,  func2, },
 { func2,  func1, }
};

int main()
{
    int i;
    printf("func1 = %p\n", func1);
    printf("func2 = %p\n", func2);

    for(i = 0; i < 2; i++)
    {
        printf("%p\n", myArray[0][i]);
        printf("%p\n", myArray[1][i]);
    }

    return 0;
}

edited Jan 9, 2014 at 13:09

answered Jan 9, 2014 at 12:24

Devolus

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3 Comments

Barmar Over a year ago

His question isn't how to use the arrays he has declared. He wants to make a new array whose elements are Read_AV_Props and Read_BV_Props.

Ginu Jacob Over a year ago

My problem is not exactly with the function pointer array. I would like to have an array of array of function pointers..

Devolus Over a year ago

Sorry, I missunderstood the question then. I retract my answer.

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