Segmentation Fault in typecasting from unsigned int to char pointer

Question

I'm new to c. Just learning. What i am trying to do is initialize a variable type of unsigned int and then store address of an char type array in it and after that I am trying to use int as a pointer and print the array using the unsigned int but I'am getting segmentation fault, I don't know why? It's printing value of memory address of char type array. But not the actual values at memory addresses. Can someone please help?

#include <stdio.h>

int main()
{
    char char_arr[4] = {'a', 'b', 'c', 'd'};
    int i;

    unsigned int hackyPointer;
    hackyPointer = (unsigned int) char_arr;

    for (i = 0; i < 4; i++)
    {
        printf("[hacky Pointer] now points to %p which contains value %c\n",hackyPointer, *((char *) hackyPointer));
        hackyPointer += sizeof(char);
    }

}

Type casting can't cause segfault. Please provide minimal reproducible example. — Eugene Sh.
– Eugene Sh., Commented Aug 21, 2019 at 15:39
When size of address is not the same as sizeof int, then there will be an error in run-time. And if you turn warnings of your compiler on, it will tell you. — Paul Ogilvie
– Paul Ogilvie, Commented Aug 21, 2019 at 15:40
I have added source.. Please check again.. I'm using sizeof.. — Virchual Shado
– Virchual Shado, Commented Aug 21, 2019 at 15:43
One obvious issue that can cause undefined behavior is that you are using %p to print an integer. The other is that your pointer size might be larger than can fit into unsigned int. — Eugene Sh.
– Eugene Sh., Commented Aug 21, 2019 at 15:46
Why are you trying to store pointer values in unsigned integer variables? Why not store pointer values in pointer variables? You're bound to have less trouble... — Steve Summit
– Steve Summit, Commented Aug 21, 2019 at 15:58

winidis · Accepted Answer · 2019-08-21 16:16:29Z

2

The problem is that you are trying to fit a bigger number than int can store, that is if you are running x64.

In 64 bit systems, pointers have a size of 8 bytes and int is usually 4 bytes. Pointer addresses work like numbers so when you cast the pointer to int, the address will be truncated.

Casting the int back to a pointer it will now contain some other address that does not belong to your program which can result in a access violation, or in other words, a segmentation fault.

Use a bigger int to hold the address, cast it to an int pointer or use uintptr_t type which is guaranteed to be big enough to hold the address.

answered Aug 21, 2019 at 16:16

winidis

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Bodo · Accepted Answer · 2019-08-21 15:52:10Z

1

What you are trying to do is the following

#include <stdio.h>
#include <stdint.h>

int main()
{
    char char_arr[4] = {'a', 'b', 'c', 'd'};
    int i;

    uintptr_t hackyPointer;
    hackyPointer = (uintptr_t) char_arr;

        for (i = 0; i < 4; i++)
        {
            printf("[hacky Pointer] now points to %p which contains value %c\n",( void * )hackyPointer, *((char *) hackyPointer));
            hackyPointer += sizeof(char);
        }
}

The program output is

[hacky Pointer] now points to 0x7ffff86d4224 which contains value a
[hacky Pointer] now points to 0x7ffff86d4225 which contains value b
[hacky Pointer] now points to 0x7ffff86d4226 which contains value c
[hacky Pointer] now points to 0x7ffff86d4227 which contains value d

You have to use an integer type that is able to contain values of pointers.

edited Aug 21, 2019 at 15:52

Bodo

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answered Aug 21, 2019 at 15:49

Vlad from Moscow

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1 Comment

Virchual Shado Over a year ago

Thanks for your great suggestion.. I got it already it worked with unsigned long int.

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