CODEWITHSUNDEEP
carrayspointers
Subash
Subash

Reputation: 43

Pointers with arrays in C

Code 1:

#include<stdio.h>
int main(void)
{
    int* arr[5];
    for(int i=0; i<5; i++) {
        arr[i] = (int*)malloc(sizeof(int));
        *arr[i] = i;
    }
    printf("%d",arr[3]);
    return 0;
}

Output:

13509232

Code 2:

#include<stdio.h>
int main(void)
{
    int* arr[5];
    for(int i=0; i<5; i++) {
        arr[i] = (int*)malloc(sizeof(int));
        *arr[i] = i;
    }
    printf("%d",arr[3][0]);
    return 0;
}

Output:

3

I've declared an array of pointers. Instead of going for *arr[3] to dereference the value stored, I can simply write arr[3][0] to dereference it. It works for all dimensions of arrays. Just by adding an extra zero to it's dimension, dereferences it's value stored at the location.

I couldn't get it, because at the very first place, arr[3] is 1D array and arr[3][0] is 2D array. After all, arr[3] is an address, and arr[3][0] is a value stored at the address arr[3]. Can I use in this way to get rid of * operator? Kindly explain in simple terms.

Thanks in advance.

EDIT 1: Let's take the simplest case, single variable.

#include<stdio.h>
int main(void)
{
    int k=5;
    int *ptr=&k;
    printf("%d",ptr[0]);
    return 0;
}

Output:

5

Upvotes: 0

Views: 89

Answers (5)

user9925578
user9925578

Reputation:

Use this type of code. And try to change you're code to this way, Hope this might help you. Cheers.

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

int main()
{
    float *p, tot = 0;
    int i, n;

    printf("Number of observations: ");
    scanf("%d", &n);

    p = (float*)malloc(n*sizeof(float));

    if(p==NULL)
    {
        printf("Memory is not allocated");
        exit(1); 
    }

    for(i = 0; i < n; i++)
    {
        scanf("%f", p+i);
    }


    for(i = 0; i < n; i++)
    {
        tot += *(p+i);
    }

    printf("\nAverage = %.2f\n", tot/n);


    return 0;
}

Sample Input:

Number of observations: 4

12.12

34.14

43.1

45.87

Sample Output:

Average = 33.81

Upvotes: -1

Afshin
Afshin

Reputation: 9173

There is an easy interpretation for using [0] to deference(although I won't recommend it). See this small code:

int i;
int j[1];

What is difference between these 2? obviously one is array with size 1, but if you think about it, array of size 1 is somehow equal to defining that type once.

dereferencing with * is somehow equal to accessing with array index. But because the data looks like to an array with 1 element, you access it with [0].

Upvotes: 0

tevemadar
tevemadar

Reputation: 13205

%d expects the value itself, not a pointer to it. arr[3] is a pointer, the 4th pointer in your pointer array. arr[3][0] is the actual number. You could also write *arr[3], and you actually do, in the loop.

Pointers behave similar to arrays, after you allocate memory to int *a, you can store a value to the pointed location as *a=123 or as a[0]=123. Also if you allocate memory for 2 integers, you can access the second item as a[1] (which is simpler), but also as *(a+1)

Bottom line: in terms of int-s, you have a 2D array, just not a tightly-packed n*m-style one, but one often reffered as jagged (https://en.wikipedia.org/wiki/Jagged_array).

Upvotes: 1

PhilMasteG
PhilMasteG

Reputation: 3185

ptr[x] can be written as *(ptr+x). So, ptr[0] translates to *(ptr+0) which is *ptr.

Upvotes: 3

Cr6
Cr6

Reputation: 1

Since the address calculated for both the cases will be same so it it fine to use they arr[3][0]. Let assume address of arr[3] to be 1000. Son address for arr[3][0] will be 1000+0*sizeof(int) i.e 1000

Upvotes: 0

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