Two's Complement Calculator — signed binary conversion

Signed and unsigned interpretation at any register width, with overflow and carry flags.

Example: -42 as a signed 8-bit value is -42 — stored as 0xD6, or 214 read unsigned.

Inputs

Value

Try a negative decimal, or a hex value with the top bit set like 0xFF.

Result

Signed value

-42

Unsigned value: 214
Hexadecimal: 0xD6
Binary: 1101 0110
Sign bit: 1 (negative)
Minimum: -128
Maximum: 127

A signed 8-bit value spans -128 to 127 — one more negative than positive. There is no positive counterpart to -128, so negating it overflows and returns -128 again.
The top bit is set, so this reads as -42 signed but 214 unsigned. The bits are identical — only the interpretation differs, which is why mismatched signedness across an interface produces nonsense.

Signed 8-bit boundaries

Value	Hex	Binary
127	0x7F	0111 1111
1	0x01	0000 0001
0	0x00	0000 0000
-1	0xFF	1111 1111
-128	0x80	1000 0000

/* generated by urdb.io/tools/twos-complement */
int8_t  as_signed   = -42;
uint8_t as_unsigned = 214U;   /* 0xD6 */

/* Same bits, different interpretation: */
assert((uint8_t)as_signed == as_unsigned);

Check it against real silicon

Chiprun runs your firmware on an emulated microcontroller and hands back the UART output, so you can confirm these numbers rather than trusting them.

Chiprun docs

Formula

negate: invert every bit, then add one

Worked example

-42 as a signed 8-bit value is -42 — stored as 0xD6, or 214 read unsigned.

start with the magnitude
42 in binary
0010 1010
invert every bit
one's complement
1101 0101
add one
1101 0101 + 1
1101 0110
That is two's complement — and it is why addition and subtraction use the same hardware regardless of sign.

Frequently asked questions

How do I convert a negative number to two's complement?

Write the magnitude in binary, invert every bit, then add one. For −42 in 8 bits: 42 is 00101010, inverted is 11010101, plus one is 11010110 — which is 0xD6, or 214 read as unsigned.

Why use two's complement rather than a sign bit?

Because addition and subtraction then work identically whether the operands are positive or negative — the same adder handles both, with no special cases. It also gives exactly one representation of zero, unlike sign-magnitude which has +0 and −0.

Why is the negative range one larger than the positive?

Because zero takes up one of the positive slots. An 8-bit signed value spans −128 to +127: 128 negative values, zero, and 127 positive. The asymmetry means −128 has no positive counterpart, so negating it overflows.

What happens if I negate the most negative value?

You get the same value back. Negating −128 in 8-bit arithmetic gives −128, because +128 is not representable. In C this is undefined behaviour for signed types, and it is a classic bug in hand-written abs() implementations.

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