Here's the work I've done so far:
2^32 = 10...0
subtract 1 -> 01...1 That gives you 32 1's
1 = 01
flip bits -> 11....10 add 1-> 1...1 That gives you 32 1's
Am I missing something? I'm asking because I tried a practice problem where you had to flip all the binary bits of a number. XOR'ing with -1 didn't work, but XOR'ing with 2^32 - 1 did.
To create a any number in 2's complement form you flip the bits and add 1.
So for -1 it is the following
1 = 00000000000000000000000000000001
flip those bits
11111111111111111111111111111110
add 1 to it and you get
11111111111111111111111111111111
Now why is that -1?
Well -1 + 1 = 0.
if you add the following together you get
11111111111111111111111111111111
+00000000000000000000000000000001
you keep doing a bit carry to the next position on the left, ultimately overflowing the field and you have 0 remaining.
00000000000000000000000000000000
A 32 bit 2's complement number can only represent values in the (inclusive) range -(2^31)..(2^31)-1. Since 2^32-1 is not in that range, it stands to reason whatever bit pattern you come up for it with actually corresponds to a value that does.
Let's number the 32 digits of an int, from 0 to 31. Digit #0 represents 2⁰, digit #1 represents 2¹, etc. More generally, digit #n represents 2ⁿ.
But we only have digits up to #31: we have 32 digits, but we started counting at 0. There is no 32nd digit to create 2³².
So what does the computer do? It basically pretends you wrote 2³² but then drops all but the first 32 digits (digits 0-31). This is called integer overflow. That means that 2³², which "should" be 1 followed by 32 0s, is actually just 32 0s -- which, of course, equals 0. Subtract 1 from that, and you get -1.
If you're working with longs (not ints), then you have 64 bits to work with instead of 32. In that case, overflow doesn't happen at bit 32, and you'll find that 2³² - 1 doesn't equal -1: it equals 4294967295, just like you'd expect. But in that case, 2⁶⁴ is 0 for similar reasons.
