How can I copy value in a struct that is passed by reference to the function and member A is also a pointer?

Question

I am learning the pointer and I stuck on an issue as I cannot debug the program because it gives me address rather than the actual array.

I am trying to create an Abstract datatype of Array

struct ARRAY_ADT
{
    int *A; // For storing the pointer to base address of the array
    int size; //For storing the size of the array
    int len; //For storing the length of the array
};

Now to insert a number in it's correct index I had written a function called Insert. I am calling it from main.

It's look like this

void insert(struct ARRAY_ADT *Array)
{
    int num;
    int index;
    int i = Array -> len - 1;

    printf("Enter the value to be inserted: ");
    scanf("%d", &num);

    printf("Enter the value to be index: ");
    scanf("%d", &index);

    printf("Array length:%d \n", Array ->len);


    if(OutofRange(Array -> size, Array -> len))
    {
        printf("Array is full:");
    }
    else
    {
        if (OutofRange(Array -> size, index))
        {
            printf("Index is not valid");
        }
        else
        {
            //int *temp = NULL;

            while(i >= index)
            {
                Array->(A+i+1) = Array->(A+i); // Problem
                i--;
            }
           
            Array -> A[index] = num;
            Array -> len++;
        }

    }
}

So, how can I copy value from right to left to make place for the new number.

How can I copy value in a struct that is passed by reference to the function and member A is also a pointer?

I just want to perform this operation

Array.A[i++] = Array.A[i];

C is not C++ and C++ is not C. Please only tag the language you are actually using. Looks like C — 463035818_is_not_an_ai, Jan 20 '21 at 16:41
`A[i++] = A[i];` is pretty much the same as `A[i] = A[i]; i++;` which is pretty much the same as `i++;`. — Ian Abbott, Jan 20 '21 at 16:43
what is the problem? `Array.A[i++] = Array.A[i];` or `Array->(A+i+1) = Array->(A+i);`? — apple apple, Jan 20 '21 at 16:44
obviously you do know the correct syntax at `Array -> A[index] = num;`? (despite the (imho) weird spacing) — apple apple, Jan 20 '21 at 16:45
@largest_prime_is_463035818 See https://gist.github.com/euppal/b473150426252dff15dc032ff5858ef3 — Import Accelerate, Jan 20 '21 at 17:09
@IanAbbott `A[i++] = A[i]` causes undefined behavior because it modifies and reads `i` between sequence points. Also, they don't want to increment `i`, because they're decrementing it on the next line. — Barmar, Jan 20 '21 at 17:36
@appleapple `Array->(A+i+1) = Array->(A+i)` is plain wrong stuff. see answer — Алексей Неудачин, Jan 20 '21 at 17:54
Thanks @Barmar it totally worked. Well I know C and C++ are different in many aspect but they both use pointer that's why I added that. However, I removed the C++ tag. I want to understand in depth why A[i++] didn't worked. Thank you everyone for all the replies. — Tarster, Jan 20 '21 at 18:14
@Barmar `A[i++] = A[i]` just does nothing . it's not ub unless out of bounds — Алексей Неудачин, Jan 20 '21 at 18:14
Thank you everyone for quick reply. Hope you all have a great day. I want to learn why A[i++] didn't work when I tried that in detail. It works in normal array but not in this. just why @appleapple. — Tarster, Jan 20 '21 at 18:16
`++i` will work. i mean `k[++i] = k[i - 1]`. `i++` will work also `k[i++ + 1] = k[i]` — Алексей Неудачин, Jan 20 '21 at 18:17
@АлексейНеудачин See https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points — Barmar, Jan 20 '21 at 18:28
@Barmar yeah. see c++11 section https://stackoverflow.com/a/4183735/7172363 — Алексей Неудачин, Jan 20 '21 at 18:44
@АлексейНеудачин Well I ask OP what is the question, I never imply it (OP's code) is correct (and in fact my [next comment](https://stackoverflow.com/questions/65813580/how-can-i-copy-value-in-a-struct-that-is-passed-by-reference-to-the-function-and?noredirect=1#comment116363754_65813580) say it's not) — apple apple, Jan 21 '21 at 06:35
@appleapple so you just told him that correct syntax of pointer notation is an array noration. correct syntax of `Array->(A+i+1) = Array->(A+i);` is `*(Array->A + i + 1) = *(Array->A + i)` . — Алексей Неудачин, Jan 21 '21 at 13:28
@АлексейНеудачин can you explicitly point out which sentence I said was wrong? Not sure what you're arguing. — apple apple, Jan 21 '21 at 14:41
@appleapple you'd offered him line of code that you said is wrong and to let him know this you'd then offered him to change notation. is it right or wrong haha ? — Алексей Неудачин, Jan 21 '21 at 16:24

score 1 · Accepted Answer · answered Jan 20 '21 at 17:42

1

Use Array->A[i+1] = Array->A[i]. Or you can shift without a loop by using memmove()

memmove(&(Array->A[i+1]), &(Array->A[i]), sizeof(Array->A[i]) * (Array->len - index));

answered Jan 20 '21 at 17:42

Barmar

741,623
53
500
612

score -1 · Answer 2 · answered Jan 20 '21 at 17:04

-1

right is

struct xntype
{
  int *b;
} xn;
*(xn.b + 1) = 15;

answered Jan 20 '21 at 17:04

Алексей Неудачин

405
4
13

How can I copy value in a struct that is passed by reference to the function and member A is also a pointer?

2 Answers2