Confusion about assigning value to a pointer with offset in C

Question

I was manually assign integer value to a pointer with offset, it appears that

ptr+1

was not actually pointed at

second_int

, or it was because the

printf

function didn't know the size of

ptr+1

?

#include <stdio.h>



int main(void)

{

    int first_int = 42;

    int second_int = 23;



    int *ptr = &first_int;

    *(ptr + 1) = second_int;



    int i = 0;

    for (i = 0; i < 2; i++) {

        printf("%d\n", *(ptr+i));

    }



    return 0;

}

And the output is:

42

1

Shouldn't it be 42 and 23?

And I found that each time I print the

*(ptr+1)

, it increments by 1.

#include <stdio.h>



int main(void)

{

    int first_int = 42;

    int second_int = 23;



    int *ptr = &first_int;

    *(ptr + 1) = second_int;



    int i = 0;

    for (i = 0; i < 10; i++) {

        printf("%d\n", *(ptr+1));

    }



    return 0;

}

And the output is:

What happened?

EDIT:

I'm only allowed to use pointer but not array (doing Learn C the Hard Way excercise). So I manually allocated memory for the

ptr

by

int *ptr = malloc(sizeof(int) * 2);

, then I assign value to

ptr

and

ptr + 1

, and the code run as expected.

Answer 1

In your code, ptr+1 is not memory managed by you.
It could be used by another program, or for any part or your program. What happens here is that the compiler optimizes away the second_int, as only its value is used. So no memory is allocated for it.
The next int in memory is fortunately also managed by you, and it is the index of your loop.
You thought memory layout was like this:
first_int | second_int | i

But it's like so:
first_int | i

If you want to be able to refer to both int with a pointer + offset, then you have to declare them as an array, to make sure that they are contiguous in memory:
int my_ints[2] = { 42, 0};
int *ptr = my_ints;
*(ptr+1) = second_int;
Then your code will work as intended.

Confusion about assigning value to a pointer with offset in C

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