Pointer to Array confusion

Question

By definition, it is a pointer variable that points to an array.

my code print the values of three element array . my question why the result is right usingprintf("Value at %p = %d\n", ptr2arr +i, *(ptr2arr[0]+i)); and wrong result except the first value while using printf("value at %p =%d\n" ,ptr2arr+i,*(ptr2arr[i]))

#include <stdio.h>
int main(int argc, char* argv[])
{
    int arr[3] = {1,2,3};
    int (*ptr2arr)[3];
    int i;
    ptr2arr = &arr;
    for(i = 0; i<3; i++)
    {
    printf("Value at %p = %d\n", ptr2arr +i, *(ptr2arr[0]+i));
    }
    printf("-------------------\n");
    for(i = 0; i<3; i++)
    {
    printf("value at %p =%d\n" ,ptr2arr+i,*(ptr2arr[i]));
    }
   return 0;
}

`

I get a compilation error for ptr2arr = &arr;. cannot convert 'int (*)[4]' to 'int (*)[3]' in assignment ptr2arr = &arr; . — chmike
– chmike, Commented Mar 6, 2015 at 15:16
@chmike works with only a warning in C, in C++ you need to match the type. I guess it's just a typo, int arr[4] should be int arr[3]. — vsoftco
– vsoftco, Commented Mar 6, 2015 at 15:19
you seem to be thinking that array[index] is the same thing as array + index. it's not. rather, it's the same thing as *(array + index). — The Paramagnetic Croissant
– The Paramagnetic Croissant, Commented Mar 6, 2015 at 15:52
The %p specifier requires an argument of type void*. If you're printing a pointer of another type, you should cast it. You can probably get away without the cast on most systems, but there's no guarantee that void* and other pointer types have the same representation. — Keith Thompson
– Keith Thompson, Commented Mar 6, 2015 at 20:23

Community · Accepted Answer · 2017-05-23 11:50:52Z

2

The expression ptr2arr[0] is the same as *ptr2arr, so it dereferences the pointer to array of 3 ints, giving you effectively the array arr. Therefore, *(ptr2arr[0] + i) is the same as *(*ptr2arr + i) is the same as *(arr + i), which gives you the correct result.

Whereas in the line

printf("value at %p =%d\n" ,ptr2arr+i,*(ptr2arr[i]));

ptr2arr[i] (syntactic sugar for ptr2arr + i) "jumps" over arrays of 3 ints, so dereferencing it *(ptr2arr[i]) gives the arr[0] only when i = 0, otherwise it gives you what's located at the address arr + 3*sizeof i (undefined behaviour).

PS: the address passed to printf should be *ptr2arr + i, not ptr2arr + i.

See also dereferencing pointer to integer array for a bit more details.

edited May 23, 2017 at 11:50

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answered Mar 6, 2015 at 15:06

vsoftco

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1 Comment

Elhaw Over a year ago

in your 3rd line answer arr+i is equivalent to &[i] in this example not arr[i]

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