Failure to use pointer array in C

Question

I'm trying to create an array of pointers which is to be used as an array of different data types.

Here, I've created the array arr to hold 5 pointers, each of which points to a different variable. a is an int, b is a float, c is a char, d is a string, fun is a function.

#include <stdio.h>
#include <float.h>
#include <string.h>

void fun(){
    printf("this is a test function");
}

int main() {
    int *arr[5];
    int a = 10;
    float b = 3.14;
    char c = 'z';
    char d[] = "hello";
    arr[0] = &a;
    arr[1] = &b;
    arr[2] = &c;
    arr[3] = &d;
    arr[4] = &fun;

    printf("a : %d\n", *arr[0]);
    printf("b : %d\n", *arr[1]);
    printf("c : %d\n", *arr[2]);
    printf("d : %d\n", *arr[3]);
    *arr[4];

    return 0;
}

I'm trying to get it to print out the values of a, b, c, and d, then execute the function fun. However, I'm getting errors like:

warning: assignment from incompatible pointer type [-Wincompatible-pointer-types]
     arr[1] = &b;

for arr[1], arr[2], arr[3], and arr[4].

Clarification: I'm trying to make an array (or, after all the comments and answers so far, an object type like struct or union) able to hold objects of different data types, including variables, functions, structs, etc. The data types are assumed to be known. I simply want to store different data types and use them in 1 object, like an array.

Comments are not for extended discussion; this conversation has been moved to chat. — Samuel Liew
– Samuel Liew, Commented Jun 14, 2019 at 8:47

babon · Accepted Answer · 2019-06-13 07:51:42Z

2

This is what you need to do:

#include <stdio.h>
#include <float.h>
#include <string.h>

void fun(){
    printf("this is a test function");
}

int main() {
    void *arr[5];
    int a = 10;
    float b = 3.14;
    char c = 'z';
    char d[] = "hello";
    arr[0] = &a;
    arr[1] = &b;
    arr[2] = &c;
    arr[3] = &d;
    arr[4] = fun;

    printf("a : %d\n", *((int *) arr[0]));
    printf("b : %f\n", *((float *) arr[1]));
    printf("c : %c\n", *((char *) arr[2]));
    printf("d : %s\n", (char *) arr[3]);

    void (*f)() = arr[4];
    f();

    return 0;
}

arr is an array of 5 pointers to void (or anything). While printing the elements, you have to cast them to the types that they actually are. For example, arr[3] is a pointer to an array (since you are taking the address of d). Since, the address of the first element is same as the start of the array, we can cast it to a pointer to a char before printing.

For the function (note that just using fun will give you the address of the function), I have assigned the address to f which is a pointer to a function accepting an unspecified arg list and returning void. Then you can invoke f().

That's the gist. Read a good book for more information.

edited Jun 13, 2019 at 7:51

answered Jun 13, 2019 at 6:56

babon

3,7842 gold badges24 silver badges24 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

ChocolateOverflow Over a year ago

What do (int *), (float *), (char *), and (char **) do here exactly? Also, why use char *d instead of char d[]?

M.M Over a year ago

arr[4] = fun; is not allowed in Standard C . Allowing conversion between void * values and function pointers (either direction) is a common but non-standard extension.

babon Over a year ago

@M.M are you saying I should use a cast like this: arr[4] = (void *) fun;?

M.M Over a year ago

@babon I'm saying the code might be rejected no matter what you do, but the reader should be aware that, if it is accepted by the compiler, it is non-standard behaviour that might not work on other compilers

babon Over a year ago

@M.M Then what is the right way to assign a function to a pointer to a void? Or are you saying there is no right way to do this?

|

Mad Physicist · Accepted Answer · 2019-06-13 07:24:25Z

1

This could be a possible use-case for a union, which would allow you to store pointers to different types:

#include <stdio.h>
#include <float.h>
#include <string.h>

void fun(){
    printf("this is a test function");
}

union u {
    int *i;
    float *f;
    char *c;
    char *s;
    void (*fn)();
};

int main() {
    union u arr[5];
    int a = 10;
    float b = 3.14;
    char c = 'z';
    char d[] = "hello";
    arr[0].i = &a;
    arr[1].f = &b;
    arr[2].c = &c;
    arr[3].s = &d;
    arr[4].fn = &fun;

    printf("a : %d\n", *arr[0].i);
    printf("b : %f\n", *arr[1].f);
    printf("c : %c\n", *arr[2].c);
    printf("d : %s\n", arr[3].s);
    (*arr[4].fn)();

    return 0;
}

You still need to keep track of which element contains what actual type, but at least the casting is not necessary, the type is actually guaranteed to be able to hold anything you want to put in it, and the fields help keep track of what you're doing.

edited Jun 13, 2019 at 7:24

answered Jun 13, 2019 at 7:16

Mad Physicist

116k29 gold badges202 silver badges292 bronze badges

2 Comments

Yunnosch Over a year ago

This is almost an answer to the problem I described in my comment. It lacks to solve the problem of knowing the type in a function which only gets the array index. I think OP needs to clarify whether I am guessing the actual problem correctly.

Yunnosch Over a year ago

Good explanation at the end. That is the problem I am referring to. Still needs clarification by OP.

Raja Sharma · Accepted Answer · 2019-06-13 06:50:03Z

0

You can change int *arr[] to void *arr[], It will work at the time of assignment level, but when you are printing you cannot cast void to int, it will throw error.

answered Jun 13, 2019 at 6:50

Raja Sharma

5201 gold badge7 silver badges20 bronze badges

1 Comment

ChocolateOverflow Over a year ago

I've also tried using void *arr[] but I'm given warning: dereferencing ‘void *’ pointer

Saurav Rai · Accepted Answer · 2019-06-13 06:50:57Z

-1

You cannot assign an integer pointer to the address of a variable which is other than an integer data type.

You can change the datatype of all the variables to integer datatype or use a generic pointer for working.

edited Jun 13, 2019 at 6:50

answered Jun 13, 2019 at 6:38

Saurav Rai

2,1971 gold badge17 silver badges31 bronze badges

2 Comments

klutt Over a year ago

"You cannot assign an integer pointer to the address of a variable which is other than an integer data type. " - Yes, you can.

Saurav Rai Over a year ago

@ John Zhau By generic pointer, I mean void pointer. link text

Collectives™ on Stack Overflow

Failure to use pointer array in C

4 Answers 4

7 Comments

2 Comments

1 Comment

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

7 Comments

2 Comments

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related