For the code below, what are the first 8 addresses accessed?
char- 1 byte, short - 2 bytes, int - 4 bytes, double - 8 bytes

double foo_arr[96]; // starts at address 0x10010000

for (int i = 0; i < 48; i++) {
foo_arr[i] = 8;
}

so the first access will be 0x10010000

then the next access will be 0x10010000 + 8 bytes, right?

I forgot how to do that in hex.

Ok, disregard that. I remember how to increment the addresses now.

however, in this problem why isn't 0x1001000 simply being incremented by 8 bytes each time?

double foo_arr[80]; // starts at address 0x10010000

for (int i = 0; i < 48; i++) {
foo_arr[i] /= 3;
}

This post was edited by Aimed_Shot on Apr 24 2013 05:54pm

lets start off with this.

what is happening

and

what do you think should happen. give examples

This post was edited by AbDuCt on Apr 24 2013 09:26pm

What do you mean? What's not being incremented? You don't have a pointer that you are incrementing, foor_arr will always point to 0x1001000 because why wouldn't it? foo_arr[1] WILL resolve to 0x1001008 though, assuming your doubles are 8 bytes.

a += b; increments the value.
a |= b; is a logical or which results in something along the lines of : 0x0001 + 0x0100 = 0x0101
a &= ~b; disables b in a, or as example: 0x1010 - 0x1000 = 0x0010

Not sure what you mean with /= - never used that one, but i think you're trying to simply do a += ?

What "are" you trying to do?

Can't you just realloc() 'x' bytes?

why is this thread resurrected