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Bitwise calculator (AND, OR, XOR, NOT and shifts)

Bitwise AND, OR, XOR, NOT, NAND, NOR, shifts and rotates on 8- to 64-bit integers, shown in decimal, hex and binary with a bit-by-bit grid.

更新日 検証済みの例：8

Decimal, or prefix with 0x, 0b or 0o. Negative numbers are stored in two's complement.
試す
結果
結果: 72
整数；最も近い値へ、等距離ならゼロから遠い値へ
Hex
0x48
Binary
0100 1000
Octal
0o110
Bits set to 1
2

A AND B is 0x48 (0100 1000), which is 72 whether read as signed or unsigned; 2 of the 8 bits are 1.

8-bit AND

A
1
1
0
0
1
0
1
0
B
0
1
0
1
1
1
0
0
Result (AND)
0
1
0
0
1
0
0
0
計算方法 S

  1. Write the inputs as 8-bit patterns

    A=1100 1010,B=0101 1100A = \texttt{1100\,1010},\quad B = \texttt{0101\,1100}

    Negative inputs are stored as 2⁸ = 256 minus their magnitude (two's complement).

  2. Apply AND bit by bit

    1100 1010∧ 0101 11000100 1000\begin{array}{rr}& \texttt{1100\,1010} \\ \land\ & \texttt{0101\,1100} \\ \hline & \texttt{0100\,1000}\end{array}

    Result bit is 1 only where both bits are 1.

  3. Read as unsigned

    ∑bi 2i=72\sum b_i\,2^i = 72

Bitwise calculator (AND, OR, XOR, NOT and shifts)について

Bitwise operators work on each bit position separately: AND gives 1 only where both bits are 1, OR where either is, XOR where they differ, and NOT flips every bit. Shifts move the whole pattern left or right, filling with zeros, or with copies of the sign bit for an arithmetic right shift, and rotates carry the bits that fall off one end back in at the other. Every result is cut to the chosen width of 8, 16, 32 or 64 bits.

Programmers use it to build and test bit masks, flags, permissions and hardware registers. The default, 0xCA AND 0x5C in 8 bits, keeps only the bits the two share: 1100 1010 AND 0101 1100 = 0100 1000, which is 0x48 or 72. XOR of the same pair gives 0x96, which reads as 150 unsigned or −106 signed.

Negative inputs are stored in two's complement, so −1 in 16 bits is 0xFFFF.

計算例

0xCA AND 0x5C (8-bit)

Operation
A AND B
A
0xCA
B
0x5C
幅
8-bit
Read the result as
Unsigned
結果
72
Hex
0x48
Binary
0100 1000
Bits set to 1
2

照合元：Python 3.8: 0xCA & 0x5C = 72 = 0x48, bin(72).count('1') = 2

XOR read as signed

Operation
A XOR B
A
0xCA
B
0x5C
幅
8-bit
Read the result as
Signed
結果
-106
Hex
0x96
Same bits, other reading
150 as unsigned

照合元：Python 3.8: 0xCA ^ 0x5C = 150; int.from_bytes(bytes([150]), 'big', signed=True) = -106

NOT 0 in 64 bits (edge)

Operation
NOT A
A
0
幅
64-bit
Read the result as
Unsigned
結果
18446744073709551615
Hex
0xFFFFFFFFFFFFFFFF
Same bits, other reading
-1 as signed
Bits set to 1
64

照合元：Python 3.8: ~0 & (2**64 - 1) = 18446744073709551615

Arithmetic shift right keeps the sign

Operation
Shift A right (arithmetic, copies the sign bit)
A
-128
Shift by (bits)
2
幅
8-bit
Read the result as
Signed
結果
-32
Hex
0xE0

照合元：Python 3.8: -128 >> 2 = -32 (Python's >> on ints is arithmetic); -32 & 0xFF = 0xE0

よくある質問

What does XOR do?

XOR (exclusive or) gives 1 where the two bits differ and 0 where they match, so 0xCA XOR 0x5C = 0x96. Two properties make it useful: x XOR x = 0, and applying the same key twice restores the original. That is why XOR toggles flags, computes parity and checksums, and appears in simple ciphers. C, Java, JavaScript and Python all write it as ^.

What is the difference between a logical and an arithmetic right shift?

A logical right shift fills the vacated top bits with 0, while an arithmetic shift copies the sign bit so negative numbers stay negative. In 8 bits, 0x80 shifted right by 2 logically is 0x20 (32), but −128 shifted right by 2 arithmetically is 0xE0 (−32), which is −128 ÷ 4. Java and JavaScript write >>> for logical and >> for arithmetic; in C, >> on a negative signed value is implementation-defined.

How does a bit mask work?

A mask is a number whose 1 bits pick the positions you care about. AND with the mask keeps those bits and clears the rest, OR sets them, XOR toggles them, and AND with NOT mask clears them. x AND 0x0F keeps the low 4 bits, and −1 AND 0x0F0F in 16 bits gives 0x0F0F = 3855, because −1 is all ones. To test bit n, check whether x AND (1 << n) is non-zero.

What does a left shift do to a number?

Shifting left by n multiplies an unsigned value by 2^n as long as no 1 bits are pushed past the top: 0x0F << 4 is 0xF0 = 240. Bits that move beyond the width are lost, so in 8 bits 0xFF << 4 is also 0xF0, not 4,080. In C, shifting by the full width or more is undefined behavior, and x86 masks the shift count to 5 bits for 32-bit operands and 6 bits for 64-bit ones.

What is popcount?

Popcount (population count, or Hamming weight) is the number of 1 bits in a value: 0x48 = 0100 1000 has 2, and NOT 0 in 64 bits has 64. x86 has a single POPCNT instruction for it, C++20 exposes it as std::popcount and Java as Integer.bitCount. It counts set flags and gives the Hamming distance between two values as popcount(a XOR b).

「Bitwise calculator (AND, OR, XOR, NOT and shifts)」の精度はどのくらいですか？

精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。 独立した出典の解答と照合した計算例：8。 例えば、「0xCA AND 0x5C (8-bit)」はPython 3.8: 0xCA & 0x5C = 72 = 0x48, bin(72).count('1') = 2と照合しています。

この計算方法の出典は何ですか？

ISO/IEC 9899:2024 (C23) §6.5.7 bitwise shift operators and §6.5.10–6.5.12 bitwise AND, XOR, OR (draft N3096); Intel 64 and IA-32 Architectures Software Developer's Manual, Vol. 2 — SAL/SAR/SHL/SHR and ROL/ROR.

この計算機について

unsigned=∑i=0w−1bi 2isigned=−bw−1 2w−1+∑i=0w−2bi 2i\begin{aligned} \text{unsigned} &= \sum_{i=0}^{w-1} b_i\,2^i \\ \text{signed} &= -b_{w-1}\,2^{w-1} + \sum_{i=0}^{w-2} b_i\,2^i\end{aligned}

出典

  1. ISO/IEC 9899:2024 (C23) §6.5.7 bitwise shift operators and §6.5.10–6.5.12 bitwise AND, XOR, OR (draft N3096)
  2. Intel 64 and IA-32 Architectures Software Developer's Manual, Vol. 2 — SAL/SAR/SHL/SHR and ROL/ROR

出典と照合済み

この計算機には、独立した出典の解答を使った計算例が 8 件あります。テストに組み込まれており、ここでも実行できます。

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