Using Pointer on 2D Array to Function

Question

I'm very new to programming so I'm sorry if this will sound as a bad question.

My Code:

#include <stdio.h>

void Order(int *waiterList){

    printf("Number is %d", *waiterList[0][1]);

}

int main(){

    int waiterList[3][3] = {{1,2,3},{4,5,6},{7,8,9}};

    Order(&waiterList);

    return 0;
}

This code is giving me this error:

pointer.c: In function 'Order':
pointer.c:5:39: error: subscripted value is neither array nor pointer nor vector
printf("Number is %d", *waiterList[0][1]);

pointer.c: In function '`enter code here`main':
pointer.c:13:8: warning: passing argument 1 of 'Order' from incompatible pointer type [- 
Wincompatible-pointer-types]
  Order(&waiterList);

pointer.c:3:6: note: expected 'int *' but argument is of type 'int (*)[3][3]'
void Order(int *waiterList){

Sorry I've never used array before but our final project is requiring us to use array and I'm finding it difficult to understand the articles I found on google... Hope you could help a student out, It's also my first time posting here because I'm kinda desperate.

Do you know how I could fix this? I would really appreciate it. Thank you! — Vihaan Khatri
– Vihaan Khatri, Commented Dec 2, 2019 at 15:08
It's simple: void Order(int waiterList[3][3]). You need not even worry about "array decay", that part is silently and implicitly handled by the compiler. — Lundin
– Lundin, Commented Dec 2, 2019 at 15:12

Sourav Ghosh · Accepted Answer · 2019-12-02 15:10:12Z

1

The simple way to achieve this would be

#include <stdio.h>

void Order(int waiterList[3][3]){  // receive an array of type `int [3][3]`
    printf("Number is %d", waiterList[0][1]);

}

int main(){

    int waiterList[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    Order(waiterList);  // just pass the array name

    return 0;
}

If you must use pointers, use proper types:

#include <stdio.h>

void Order(int (*waiterList)[3][3]){   // pointer to an array of type `int[3][3]`

    printf("Number is %d", (*waiterList)[0][1]);

}

int main(){

    int waiterList[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    Order(&waiterList);  // pass the address of the array

    return 0;
}

answered Dec 2, 2019 at 15:10

Sourav Ghosh

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1 Comment

Vihaan Khatri Over a year ago

This was really helpful. Thank you very much!

Some programmer dude · Accepted Answer · 2019-12-02 15:16:34Z

0

Arrays decay to a pointer to its first element, for a generic array a it decays to &a[0].

In the case of an arrays like your waiterList in the main function, when it decays to a pointer to its first element (&waiterList[0]) it will have the type "pointer to an array". More specifically for waiterList the type becomes int (*)[3]. Which has to be the type of the argument for the Order function:

void Order(int (*waiterList)[3]){
    printf("Number is %d", waiterList[0][1]);
}

You call it simply passing the array as any other variable, no pointer or address-of operator needed:

Order(waiterList);

edited Dec 2, 2019 at 15:16

answered Dec 2, 2019 at 15:10

Some programmer dude

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3 Comments

Sourav Ghosh Over a year ago

Did you mean (*waiterList)[0][1]?

Some programmer dude Over a year ago

@SouravGhosh No, as waiterList is a pointer then waiterList[0] dereferences the pointer already (remember it's equal to *(waiterList + 0)). That dereference will give an array which is then indexed the "normal" way.

Lundin Over a year ago

@SouravGhosh No it's correct, he doesn't use a pointer to array with two dimensions as in your example. But of course there's no need to use array pointer syntax at all here. int waiterList[3][3] is the most readable parameter declaration.

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