Re: Explain why this prints the same

In article <1133969869.613104.235330@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> "Eric Lilja" <mindcooler@xxxxxxxxx> writes:
> Hello, I'm trying to help someone on a linux-oriented forum. I've taken
> his original code and cleaned it up, but I am still wondering about
> something. Here's the code:

Ok, let's see:

> #include <stdio.h>
>
> int
> main(void)
> {
> int a[][3]={ {1,2,3},{4,5,6,},{7,8,9} };

Well, for a a segment of 9 integers is allocated succesively filled with
the integers 1 to 9.

> int *b = (int *)a;

As 'a' in this context is a pointer to an array of three ints, you are
lying to the compiler. The result is (I think) undefined behaviour.

> int **c = (int **)a;

And more so. Here the result is certain to be undefined behaviour.

> int i = 0;
> int num_elements = sizeof(a) / sizeof(int);
>
> for(i = 0; i< num_elements; ++i)
> printf("*b = %d *c = %d\n", *b++, *c++);

And this is the third time you are lying. You are lucky that
sizeof(int) == sizeof(*int) on your system.
>
> return 0;
> }
....
> The output is:
> *b = 1 *c = 1
....
> *b = 9 *c = 9
>
> Why is it same for *b and *c?

Because on your system sizeof(int) == sizeof(*int) and the compiler
disregards all those lies.
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