I'm reading The C Programming Language by K & R 2nd edition and got to the bitwise operators. Why, on C, the Complement (~) Operator applied to a N number, returns -(N + 1) ? When it normally just flips the 1 bits to 0 and the 0 bits to 1.

I'm reading The C Programming Language by K & R 2nd edition and got to the bitwise operators. Why, on C, the Complement (~) Operator applied to a N number, returns -(N + 1) ? When it normally just flips the 1 bits to 0 and the 0 bits to 1.

• c
• binary
• bit-manipulation
• twos-complement

Share Improve this question Follow edited Mar 6 at 14:11 nerdpanda asked Mar 6 at 8:57 nerdpandanerdpanda 1122 bronze badges 3

• 3 The Wikipedia article does not explain it detailed enough? – Axel Kemper Commented Mar 6 at 9:14
• Please edit and make clear what "The C Prog. Lang." is. Is it a book, then write out the full title and the author. If it's web site, show the link. – Jabberwocky Commented Mar 6 at 10:02
• For any number n in C, -n is the negative equivalent. Regardless of if 2's complement or some other signed system is used. Perhaps you mean ~N + 1? – Lundin Commented Mar 6 at 10:39

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2 Answers 2

Sorted by: Reset to default 1

Consider some bit string b, and let u be the number we obtain by interpreting b as an unsigned binary numeral. So, with bit string 10110001, u = 128 + 32 + 16 + 1 = 177.

Two’s complement interprets a fixed-length bit string of n bits:

• If the first bit is 0, the bit string represents the value u.
• If the first bit is 1, the bit string represents the value u−2ⁿ.

So, in eight-bit two’s complement, 10110001 represents 177−2⁸ = 177−256 = −79.

Now consider the ~ operator. This operator inverts each bit in a bit string. 10110001 becomes 01001110.

There is another operation that inverts each bit. If we subtract the bit string as a binary numeral from another bit string that is all 1s, of the same length, it inverts all the bits: 11111111₂ − 10110001₂ becomes 01001110₂. This is because, in the subtracting 1−x, where x is a single bit, produces 1 when x is 0 and 0 when x is 1, with no borrow.

Now observe that 11111111₂ = 100000000₂ − 1₂. In general, for n bits, a binary numeral of n 1s equals 2ⁿ − 1.

So the result of ~ applied to an unsigned value u of n bits is 2ⁿ − 1 − u.

Now let’s consider applying ~ to a signed value s of n bits:

• If s is non-negative, its bits interpreted as an unsigned binary numeral yield the same value: u = s. Then ~ produces 2ⁿ − 1 − u. That bit string starts with 1, so its interpretation in two’s complement is 2ⁿ − 1 − u − 2ⁿ = −1−u = −(u+1) = −(s+1).
• If s is negative, its two’s complement interpretation s equals u−2ⁿ. So u = s + 2ⁿ. Applying ~ to that gives us 2ⁿ − 1 − u = 2ⁿ − 1 − (s + 2ⁿ) = −1 − s = −(s+1). That bit string starts with 0, so its interpretation in two’s complement is −(s+1).

Thus, when ~ is applied to a signed value s, the result is −(s+1).

(Historical note: The 2024 version of the C standard requires two’s complement for signed integer types. Long ago, some C implementations used one’s complement or sign-and-magnitude, and this was permitted in earlier versions of the standard.)

Basically you are working with bits.

The ~ inverts the bits of a number

5 = 00000101
~5 = 11111010

The ~5 is -6 because of the form -(N+1):
~5 = -(5+1) = -6 = 11111010

In terms of the left-bit, if you have a left-bit that is a 0 it means that your N is a positive number or 0. If you have a 1 in the left-bit than it means that your N is a negative number.

Ex.:

5 = 00000101 // starts with a 0 -> positive
~5 (-6) = 11111010 // starts with a 1 -> negative