# Using ~ (Unary bitwise complement) for Zero

posted 5 years ago

I wrote a program to find 0's complement and found out that its -1.

So I get it that ~ is giving me 2's complement. But what I dont understand how is 0's 2's complement is -1.

00000000 is 0

11111111 is 1's complement

1 00000000 is 2's complement of above where carry 1 is ignored.

So why the result is -1 (11111111)

If 1's complement was used then it would give me -0 which is wrong.

So I get it that ~ is giving me 2's complement. But what I dont understand how is 0's 2's complement is -1.

00000000 is 0

11111111 is 1's complement

1 00000000 is 2's complement of above where carry 1 is ignored.

So why the result is -1 (11111111)

If 1's complement was used then it would give me -0 which is wrong.

Campbell Ritchie

Marshal

Posts: 52590

119

Stephan van Hulst

Bartender

Posts: 6584

84

posted 5 years ago

In Java's representation (which uses two's complement), yes.

*The mind is a strange and wonderful thing. I'm not sure that it will ever be able to figure itself out, everything else, maybe. From the atom to the universe, everything, except itself.*

posted 5 years ago

WRONG.

~0 is NOT two's compliment, it simply flips every '0' bit to '1', and every '1' bit to '0'.

In Java, 2's compliment for 0 is 0. In fact, in ANY language, 2's compliment of 0 is 0, as 2's compliment is an algorithm, not a language specific feature.

Pranav Raulkar wrote:So in java 2's complement for 0 is 11111111 which is -1. Agreed.

WRONG.

~0 is NOT two's compliment, it simply flips every '0' bit to '1', and every '1' bit to '0'.

In Java, 2's compliment for 0 is 0. In fact, in ANY language, 2's compliment of 0 is 0, as 2's compliment is an algorithm, not a language specific feature.

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Campbell Ritchie

Marshal

Posts: 52590

119

posted 5 years ago

This is how two’s complement numbers are worked out:You work out the range of numbers available: for 8 bits that is 256 You allow exactly half that range for negative numbers: -1 to -128 inclusive. You allow exactly the other half of the range for non-negative numbers: 0 to 127 inclusive. For non-negative numbers, use exactly the same format as for unsigned numbers, 0000_0000 to 0111_1111 inclusive. For negative numbers, subtract the absolute value from the size of the range. You can work out the two’s complement value of a negative number by subtracting its absolute value from the size of the whole range. For example: the two’s complement representation of -97 is the same as the unsigned representation of 256 - 97, or 1_0000_0000 - 0110_0001

That is how complementary numbers are really defined.

If you try that calculation, you get this for -97:

There are at least two other ways you can think of a two’s complement number (still in 8 bits):

One way: The leftmost bit (no 7) represents -2^7 (-128)or 0, the next bit represents (+)64 or 0, then (+)32 or 0, etc until the rightmost bit (the 0-th bit) represents (+)1 or 0. So 1001_1111 means -128 + 0 + 0 + 16 + 8 + 4 + 2 + 1 = -97.

The other way is to remember that 256 - 97 = 256 - 1 - 97 + 1. The -1 and +1 cancel out, but look good in binary.

256 - 1 looks like this:

Now you can subtract 97 from 255

Now you add 1 again

Did you notice that subtracting from 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 is the same as swapping all the bits? So, you can get something identical to two’s complement by swapping all the bits and adding 1. If you try complementing 0 to -1 and adding 1, you get 0. Try it

No. Not at all. I think you have misunderstood two’s complement arithmetic. The two’s complement representation of 0 is 0000_0000 (in 8 bits). The two’s complement of 0 is not 1111_1111. Never.Pranav Raulkar wrote:So in java 2's complement for 0 is 11111111 which is -1. Agreed.

This is how two’s complement numbers are worked out:

That is how complementary numbers are really defined.

If you try that calculation, you get this for -97:

`1_0000_0000`

1001_1111

__0110_0001__-1001_1111

There are at least two other ways you can think of a two’s complement number (still in 8 bits):

One way: The leftmost bit (no 7) represents -2^7 (-128)or 0, the next bit represents (+)64 or 0, then (+)32 or 0, etc until the rightmost bit (the 0-th bit) represents (+)1 or 0. So 1001_1111 means -128 + 0 + 0 + 16 + 8 + 4 + 2 + 1 = -97.

The other way is to remember that 256 - 97 = 256 - 1 - 97 + 1. The -1 and +1 cancel out, but look good in binary.

256 - 1 looks like this:

`1_0000_0000`

1111_1111. . . 255 in unsigned numbers.

__0000_0001__-1111_1111

Now you can subtract 97 from 255

`1111_1111`

1001_1110

__0110_0001__-1001_1110

Now you add 1 again

`1001_1110`

1001_1111. . . and lo and behold, we have -97

__0000_0001__+1001_1111

Did you notice that subtracting from 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 is the same as swapping all the bits? So, you can get something identical to two’s complement by swapping all the bits and adding 1. If you try complementing 0 to -1 and adding 1, you get 0. Try it

No, you are not calculating a two’s complement at all. What you are doing is taking the bit pattern, eg 0110_0001 for 97 and 0000_0000 for 0 (in 8 bits) and getting the complement of that bit pattern. That is equal to -(i + 1). As you saw above, 97 turns into -98 and 0 turns into -1.But if you calculate 2's complement it comes out to be 00000000. Correct me if I'm wrong from my first post.

posted 5 years ago
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Note: There is a difference between the "complement" and the "two's compliment". I assume you want the latter.

0 is

0000 0000 0000 0000 (i'm only going to use 16 bits, but it works the same for 32, or 64 or however many bits you want)

invert all the bits, you get

1111 1111 1111 1111

add 1:

but that left-most 1 doesn't fit. We run out of bits, so it falls off the end, leaving us with

0000 0000 0000 0000

0 is

0000 0000 0000 0000 (i'm only going to use 16 bits, but it works the same for 32, or 64 or however many bits you want)

invert all the bits, you get

1111 1111 1111 1111

add 1:

but that left-most 1 doesn't fit. We run out of bits, so it falls off the end, leaving us with

0000 0000 0000 0000

Campbell Ritchie

Marshal

Posts: 52590

119

posted 5 years ago

The wikipedia article about two’s complement is slightly imprecise. Two’s complement is made by subtracting from 2^

Fred has already done that, only for 16 bits. It is exactly the same, but occupies more space on the screen.Pranav Raulkar wrote: . . . Can you put up a claculation just like you did for -97 for complement of zero?

The wikipedia article about two’s complement is slightly imprecise. Two’s complement is made by subtracting from 2^

`i`where

`i`is the number of available bits. Inverting each bit and adding one is not how it is defined, but

__always__gives the same result, provided the numbers are within the range

`-2^(`.

*i*- 1)...2^(*i*- 1) - 1
posted 5 years ago

Agreed! Complement and 2's complement are different.

If we do complement of 0 its 11111111 which is 255.

since ~ is complement operator ~0 should give me 255. Why -1? If its giving me -1 it means ~ is 2's complement.

But we all know 2's complement for 0 is 00000000 (Carry over 1 is discarded)

I'm so LOST!

If we do complement of 0 its 11111111 which is 255.

since ~ is complement operator ~0 should give me 255. Why -1? If its giving me -1 it means ~ is 2's complement.

But we all know 2's complement for 0 is 00000000 (Carry over 1 is discarded)

I'm so LOST!

posted 5 years ago
There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Again, you're making a few faulty assumptions here.

the complement of 0 will

0000 0000 0000 0000 0000 0000 0000 0000

or

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

the complement will be (for the int)

1111 1111 1111 1111 1111 1111 1111 1111

To determine the value, you have to do the following:

look at the left most bit. If it is a zero (it our case it isn't), you simply add up the powers of 2 that correspond to the 1's, and the value is positive.

if the left-most bit IS a 1, your result will be negative. next, take the 2's complement of the number. So, we flip all the bits, and add 1. so it becomes

0000 0000 0000 0000 0000 0000 0000 0000

+ 1

=============================

0000 0000 0000 0000 0000 0000 0000 0001

So "1111 1111 1111 1111 1111 1111 1111 1111" represents "-1".

the complement of 0 will

*slightly*depend on whether you have an int or a long. an int is 32 bits, and a long is 64 bits. so, 0 will be either0000 0000 0000 0000 0000 0000 0000 0000

or

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

the complement will be (for the int)

1111 1111 1111 1111 1111 1111 1111 1111

To determine the value, you have to do the following:

look at the left most bit. If it is a zero (it our case it isn't), you simply add up the powers of 2 that correspond to the 1's, and the value is positive.

if the left-most bit IS a 1, your result will be negative. next, take the 2's complement of the number. So, we flip all the bits, and add 1. so it becomes

0000 0000 0000 0000 0000 0000 0000 0000

+ 1

=============================

0000 0000 0000 0000 0000 0000 0000 0001

So "1111 1111 1111 1111 1111 1111 1111 1111" represents "-1".

Campbell Ritchie

Marshal

Posts: 52590

119

posted 5 years ago

1111_1111 is only 255 in unsigned binary numbers. It is meaningless to say something like “1010_1010 means xyz”, without saying what format you are using, and what memory size. There are at least four formats for binary integers, possibly five if you include one’s complement. In two’s complement in eight bits, 1111_1111 means -1decimal.

Campbell Ritchie

Marshal

Posts: 52590

119

posted 5 years ago

I have already told you, and I think Fred has too, that the ~ operator does not produce the two’s complement. It returns the bit pattern inverted, which is more akin to one’s complement. So you get the one’s complement of 0 which in two’s complement returns -1decimal. Remember you had to add 1 to ~97 to get -97.

Campbell Ritchie

Marshal

Posts: 52590

119

posted 5 years ago

~~1_~~

`0000_0000`Flip the bits. What the ~ operator does, giving the result on the next line.

`1111_1111`That returns -1decimal, but to get that into two’s complement, add 1, and the 8th bit vanishes into cyber-limbo never to be seen again.

`0000_0000`Now we get back to 0. But that isn’t what the ~ operator does. It only does what you saw one line up.