Question

The following bit of code…

void foo(char* x)
{
        int i;
        int len = sizeof(x)/sizeof(x[0]);

        printf("len: %d\n", len);
        for(i=0; i<len; i++){
                printf("i: %d, v: %x\n", i, x[i]);
        }
}

…when called as:

        char bar[] = {0xDE, 0xAD, 0xBE, 0xEF};
        foo(bar);

…outputs:

len: 8
i: 0, v: ffffffde
i: 1, v: ffffffad
i: 2, v: ffffffbe
i: 3, v: ffffffef
i: 4, v: ffffffff
i: 5, v: 7f
i: 6, v: 0
i: 7, v: 0

Not exactly sure if it’s applicable to this but reading through similar posts on SO, it seems to be an alignment issue (I’m on a 64 bit machine).

Any pointers?

Thanks.

Answer 1

This has nothing do to with alignment.

int len = sizeof(x)/sizeof(x[0]);

x is a pointer and not an array. sizeof (x) computes the size of the pointer which is 8 bytes on your 64-bit machine.

Now it prints ffffffde insteas of de because of integer promotion (sign extension is performed). To avoid the ffffff you can use unsigned char type instead of char or cast the printf arguments to unsigned int like this:

printf("i: %d, v: %x\n", i, (unsigned int) x[i]);