how to call a function with a pointer as its argument

Question

what is problem? i want send a multi-dimensional pointer to a function as its arguments. But i got this error

Exception thrown at 0x00007FF770A92231 in Check nevis.exe: 0xC0000005: Access violation reading location 0x0000000400000007.

what did i do wrong? and what this error means?

#include <stdio.h>
test(int *ptr)
{
    printf("%d", *(*(ptr+1)+1));
}
int main()
{
    int a[2][3] = { { 1,2,3 }, { 4,5,6 } };
    int(*ptr)[2][3]=&a;
    test(ptr);
    return 0;
}

with an int* pointer, you can't use * twice on it! What did you want to print? — B. Go, Dec 07 '19 at 20:42
`int(*ptr)[2][3]=&a;` should most likely be just `int** ptr = a` — B. Go, Dec 07 '19 at 20:47
then your function should be on int** to access the 2 dimensions — B. Go, Dec 07 '19 at 20:47

score -1 · Answer 1 · answered Dec 07 '19 at 20:54

-1

You need to have pointer to pointer to int:

void test(int **ptr)
{
    printf("%d", ptr[1][1]);
}

int main()
{
    int a[2][3] = { { 1,2,3 }, { 4,5,6 } };
    int **ptr = a;
    test(ptr);
    return 0;
}

Note: it is expected and good practice to prefix the function, which returns nothing, with 'void'. Probably you got a warning about it - NEVER ignore warnings!

answered Dec 07 '19 at 20:54

VillageTech

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18

what is difference between ptr[1][1] and *(*(ptr+1)) – Asadiyan Dec 07 '19 at 20:57
i tried the code and i got the same error again – Asadiyan Dec 07 '19 at 21:01
and the same location? – VillageTech Dec 07 '19 at 21:10
i gotException thrown at 0x00007FF6CB422243 in Check nevis.exe: 0xC0000005: Access violation reading location 0x0000000400000007 – Asadiyan Dec 07 '19 at 21:13
Did you change the main() function as sugested, too? – VillageTech Dec 07 '19 at 21:20
no i didnt change main function – Asadiyan Dec 07 '19 at 21:25
So you have an answer. The int(*ptr)[2][3]=&a; means that you try to write address of array 'a' into element [2][3] of an unknown (unallocated) array, pointed by 'ptr'. It is wrong. Change this line as suggested. – VillageTech Dec 07 '19 at 21:28
no i want value – Asadiyan Dec 07 '19 at 21:32
What value??????????? Try do this, as suggested, and you will see that it works as you want. BTW: there is NO element [2][3] - this is indice range, it means that maximum indices are 1 and 2. – VillageTech Dec 07 '19 at 21:35
1

`int **ptr` **most certainly** is **not** compatible with `int (*)[3]` – Antti Haapala -- Слава Україні Dec 07 '19 at 22:07
2

And `*(*(ptr + 1))` is the same as `ptr[1][0]`, not the other way round. – Antti Haapala -- Слава Україні Dec 07 '19 at 22:09
Of course. Should be? Your array is an array of two 3-element subarrays, so the first '*' makes pointer to the subarray, the second '*' makes pointer to subarray element. – VillageTech Dec 07 '19 at 22:11
About indices order: you are right, my mistake. – VillageTech Dec 07 '19 at 22:13
No, it dont work i tried all of ur suggestions – Asadiyan Dec 08 '19 at 10:36

how to call a function with a pointer as its argument

1 Answers1