CODEWITHSUNDEEP
carrayspointers
Antonio MG
Antonio MG

Reputation: 20410

Array size in multidimensional array with pointers

I'm using a library that creates a multidimensional array in this way:

const unsigned short arrayA[5] = {1, 2, 3, 4, 5};
const unsigned short arrayB[3] = {7, 8, 9};

const unsigned short *multiArray[] ={arrayA, arrayB};

I get the values of it and it works:

printf("03: %d\n", multiArray[0][3]); //4
printf("12: %d\n", multiArray[1][2]); //9

The problem comes when I need to get the size of any of the arrays, I've tried this:

printf("Size of first array: %ld \n", sizeof(multiArray[0])/sizeof(multiArray[0][0]));

It returns 2, probably because it's using addresses.

How can I get the size of the array then?

I have to do this trying not to change the way the arrays are declared, so the arrays are already there and I just need to get the size.

Any approaches?

Upvotes: 2

Views: 4586

Answers (6)

edgar
edgar

Reputation: 21

It is possible to obtain both the amount of rows and columns of a matrix:

int main(void) {
     int row;
     int column;
     int matrix[4][5]={{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5}};

     row=sizeof(matrix)/sizeof(matrix[0]); 
     column=sizeof(matrix[0])/row;

     printf("row %i \n",row);
     printf("column %i \n",column);
     return 0;
}

Upvotes: 2

David Ranieri
David Ranieri

Reputation: 41046

As other have said, you can not, instead you can build your own type:

#include <stdio.h>

struct cont {
    const unsigned short *value;
    size_t len;
};

int main(void)
{
    const unsigned short arrayA[5] = {1, 2, 3, 4, 5};
    const unsigned short arrayB[3] = {7, 8, 9};

    struct cont arrays[] = {
        {arrayA, sizeof(arrayA) / sizeof(arrayA[0])},
        {arrayB, sizeof(arrayB) / sizeof(arrayB[0])},
    };

    printf("03: %d\n", arrays[0].value[3]);
    printf("12: %d\n", arrays[1].value[2]);


    printf("Size of first array: %ld \n", arrays[0].len);

    return 0;
}

Upvotes: 1

Roee Gavirel
Roee Gavirel

Reputation: 19463

You cant! you should hold this data and pass it to wherever you need.

BTW when you do this: sizeof(multiArray[0])/sizeof(multiArray[0][0])
you basically do sizeof(<pointer>)/sizeof(<unsigned short>) => 2 on your specific machine.

Upvotes: 1

Ifthikhan
Ifthikhan

Reputation: 1474

It is not possible to deduce the size of the arrays pointed to by the pointers in multiArray since there is no way of performing any computations as the two arrays (arrayA and arrayB) will be stored in arbitrary locations. A good practice is to store the size of the arrays.

Upvotes: 4

Some programmer dude
Some programmer dude

Reputation: 409442

If you do sizeof on a pointer, you get the size of the actual pointer and not what it points to. You have to keep track of the size of the contained array some other way.

Upvotes: 2

aragaer
aragaer

Reputation: 17858

You can't. You have to know the size of array in advance and pass it as additional argument.

Upvotes: 2

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