CODEWITHSUNDEEP
cmathpointers
user830090
user830090

Reputation:

C Pointer Arithmetic Problems

I'm having a little fiddle with pointer arithmetic and just pointer in general and I've pulled together this code. 

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

int main(int argc,char **argv){
    void *ptr=NULL;
    char a='a';
    int i0=0;
    char b='b';
    int i1=1;
    char c='c';
    int i2=2;
    ptr=&a;
    //What do I do here to get to i0?
    printf("%d",(int*)ptr); //and to print it out?
    while(1);
    return 0;
}

My question is exactly what I put into the comments. How do I get ptr to point to i0 without doing 'ptr=&i0;' using pointer arithmetic? Also how do I then print it out correctly for characters and integers(one method for char and one for int).

Thanks in advance. ;)

Upvotes: 0

Views: 734

Answers (8)

Per Johansson
Per Johansson

Reputation: 6877

You'll have to do something like this:

int main(void)
{
    union u
    {
       char ch;
       int i;
    } arr[] = { { .ch = 'a' }, { .i = 0 }, { .ch = 'b' }, { .i = 1 } };

    union u *ptr = &arr[0];
    printf("%d\n", ptr[1].i);
}

Not exactly what you wrote but would work.

Upvotes: 0

ciamej
ciamej

Reputation: 7068

the only thing I can think of is:

ptr += ((char*)&i0 - &a)

but not sure if it's valid anyway and it doesn't make much difference from ptr = &i0

as others pointed out, in accordance to the standard you can only do pointer arithmetic within bounds of a single object, like a struct or array, so the code above even if it might work is not really correct

Upvotes: 0

phihag
phihag

Reputation: 287755

The only way to get a pointer to the location of i0 is & i0.

While some compilers may align i0 so that * ((int*) (((char*) ptr) - 1)) can be used to access i0, that's not guaranteed. Compilers often reorder local variables or may not even store them in memory at all.

Upvotes: 7

pmg
pmg

Reputation: 108968

You can only do pointer arithmetic within a whole object (like an array). This works:

#include <stdio.h>
int main(void) {
    int arr[100];
    int *ptr;
    arr[42] = 42;
    ptr = arr; /* ptr points to the 1st element */
    printf("%d\n", *(ptr + 42)); /* contents of the 43rd element */
    return 0;
}

Upvotes: 1

nos
nos

Reputation: 229058

You can't do this.

Pointer arithmetic works on pointer to array elements that is, you can do arithmetic on a pointer that points to an element within an array to adjust the pointer to another element within that same array.

You cannot do pointer arithmetic on a pointer to a variable and make it point to another unrelated variable.

Upvotes: 1

John Humphreys
John Humphreys

Reputation: 39254

Your code doesn't make sense. You have no guarantee of where a, i0, b, i1, c and i2 are defined in memory when they're created - so you can't use pointer arithmetic to move from the address of a (ptr=&a) to i0.

If you want ptr to equal i0's location you can do ptr = &i0. If i0 was an integer (which it is) it will be 4 bytes big, so you can use pointer arithmetic to move through that integer 1 btye at a time if your pointer is void/char.

Upvotes: 1

AnT stands with Russia
AnT stands with Russia

Reputation: 320381

You can't get ptr to point to i0 using pointer arithmetic. Pointer arithmetic only works within the bounds of a single array object (non-array variables are treated as arrays of size 1). You can't use pointer arithmetic to make a pointer skip from one standalone object to another.

Upvotes: 1

Alok Save
Alok Save

Reputation: 206508

In this code sample, without doing 

ptr=&i0;

You cannot get ptr to point to i0 portably because,
i0 is just a local variable & it's address is not stored in any other variable.

Upvotes: 0

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