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When I run the following code:

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

int main(void) {
    int array[100];
    int (*array_ptr)[100];
    void *buff = malloc(500);
    int *ptr;
    printf("array: %p \narray+1: %p\n", array, array+1);
    printf("array_ptr: %p \narray_ptr+1: %p\n", array_ptr, array_ptr+1);
    printf("buff: %p\n", buff);
    printf("ptr: %p\n", ptr);
}

the result is like this:

array: 0x7fffe6dc6bd0
array+1: 0x7fffe6dc6bd4
array_ptr: (nil)
array_ptr+1: 0x190
buff: 0x1f80260
ptr: 0x7fffe6dd417c

I run it multiple times, array, array+1, buff and ptr all change values randomly, but array_ptr and array_prt+1 never change, although the pointer arithmetic result 0x190 is as expected.

Does it indicate that the array pointed by array_ptr is stored in heap? But the dynamically allocated memory chunk pointed by buff is also supposed to be in heap and its value changes, why is that? Thanks!

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  • 6
    Your pointer array_ptr is uninitialized, it doesn't point anywhere, its value will be indeterminate. Same with ptr. Commented Feb 24, 2020 at 7:28
  • And in C you shouldn't really cast the result of malloc. Commented Feb 24, 2020 at 7:30
  • Thanks @Someprogrammerdude but why is array_ptr+1's value still available? Commented Feb 24, 2020 at 7:30
  • Lastly, the only things stores in the heap is the 500 bytes you allocate with malloc. But note that the storage for the variable buff itself isn't on the heap, it only points to the heap. Commented Feb 24, 2020 at 7:31
  • 1
    This can´t get compiled. You need to #include <stdlib.h>in order to use malloc. Commented Feb 24, 2020 at 9:13

2 Answers 2

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array_ptr is uninitialized pointer, and gets an undefined value(in your case 0).

array_ptr+1 is sizeof(int)*100 = 400 = 0x190 more than array_ptr

ptr is also an uninit pointer, which in your case points to garbage.

You need to initialize pointers after they're defined to get any valid results

To your question, array is on the stack, buff is on the heap, and ptr/array_ptr are uninitialized will give you garbage or segmentation fault if you try to access their data

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Does it indicate that the array pointed by array_ptr is stored in heap?

No. array_ptr points to nothing, neither on the stack nor on the heap.

int (*array_ptr)[100];

array_ptr is one pointer to an array of 100 int objects, but with this statement you do not create an array with 100 int objects by which array_ptr is pointing to the first element of this array. It only creates the pointer itself.

With:

printf("array_ptr: %p \narray_ptr+1: %p\n", array_ptr, array_ptr+1);

You are trying to print the addresses of objects the pointer array_ptr and its offset by 1 point to, but this isn´t possible since array_ptr is not initialized to point to such objects. Thus, You´ll get any kind of Undefined Behavior/ unpredictable results.

You need to initialize array_ptr with, f.e. the address of the first int object of array:

int (*array_ptr)[100] = array;

The same goes for the pointer ptr with:

printf("ptr: %p\n", ptr);

ptr needs to have an address of an object it points to in order to show the address of that object.

But the dynamically allocated memory chunk pointed by buff is also supposed to be in heap and its value changes, why is that?

This is a completely different thing. With malloc you do allocate memory on the heap (if it was successful of course) and you do get a pointer to that memory back, which isn´t the case with int (*array_ptr)[100];.

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1 Comment

The undefined behavior in this case seems to be that the uninitialized variable array_ptr has the value 0 (whether that happens to be the incidental value on the stack where the variable resides or whether it's a debug mode initialization we don't know); and consequently, incrementing it points "to the next array", which would be 400 resp. 0x190 bytes behind it (the memory occupied by an array of 100 ints). 0 + 400 = 400, which is the address printed.

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