changing arr[0] to arr[1]

[Darkroman]

Here's what I mean

void dectobin (int * arr, unsigned char b)
{
    arr[0] = (b & 128) ? 1 : 0;
    if (b) dectobin (&arr[1], b << 1);
}

&arr[1] is what I'm mostly concerned about. I tested it out in some code:

void testarray (int arr[])
{
    arr[0] = 255;
}

// ...

for (int i = 0; i < 8; ++i)
    {
        testarray(&testarr[1]);
    }
for (int i = 0; i < 8; ++i)
    {
        cout << "[" << i << "]: " << testarr[i] << endl;
    }

Indeed it did what I thought it would. Inside the function arr[0] is actually arr[1] because &testarr[1] makes arr[0] == arr[1]... wait, what?? Can someone explain to me the behind the scenes? The best explanation I have is &testarr[1] sets the address of arr[0] as arr[1]. My brain is stuck in an infinite loop of sorts. BTW, yes I know what the main code is, and fairly understand bitwise operations and truth tables.

[Salem]

It doesn't make any difference to testarray, it's just a pointer to an int.
The fact that you made the parameter as int arr[] doesn't convey any extra arrayness properties.

You could do this.

int array[10];
int val;
int *p = malloc(10,sizeof(*p);

// Then any of these
testarray(array);
testarray(&array[0]);
testarray(&array[1]);
testarray(&val);
testarray(p);
testarray(&p[0]);
testarray(&p[1]);

Just make sure the function in question doesn't try to access beyond the bounds of the supplied parameter.