How to do bitwise operation in shell

$((...))

$ echo $(( 3 * 4 )) 12

It is especially convenient in some scenarios, which can save the trouble of writing programs, such as verifying certain operations.

The following is the scenario of verifying deep understanding of computing system exercise 2. 12.

The expression ~0 will generate a mask of all 1, which is portable regardless of the word length of the machine.

$ printf " % x \ n " $((~ 0))ffffffffffffffff

$ printf " % # x \ n " $((~ 0))0x ffffffffffffffff

The above test shows that in the shell, the bitwise inversion of 0 is 64 bits.

Print the leading characters of the shell's printf command: Table 7-4 of the shell Script Learning Guide describes the meaning of the # symbol in the format parameters, and you can use the # to output the leading characters 0x (16 hexadecimal) and 0 (octal).

X & amp0xFF produces a value consisting of the least significant byte of x.

$ printf " % # x \ n " $((0x 89 abcdef & amp； 0xFF ))0xef

$ printf " % # . 8x \ n " $((0x 89 abcdef & amp； 0xFF ))0x000000ef

The following x = 0x8765432 1

The least significant byte and all other bits of A.x are set to 0.

$ printf " % # . 8x \ n " $((0x 8765432 1 & amp； 0xFF )) 0x0000002 1

$ printf " % # . 8x \ n " $((0x 8765432 1 & amp； ? bash:0x 8765432 1 & amp； ? 0xFF: Syntax error: Expected operand (error flag is "? 0xFF”)

The answer to the exercise given in the book is "X &；; ? 0xFF "、"？ "Here you are. The number 1 verifies that the shell is not working properly.

B All positions except the least significant byte of X are complemented, and the least significant byte remains unchanged.

$ printf " % # x " $((0x 8765432 1 ^ ~ 0x ff))0x ffffffff 789 ABC 2 1

Because ~0xff generates a 64-bit mask, the result is somewhat unexpected, but the last 32 bits are expected.

The least significant byte of C.x is set to all 1, and all other bytes remain unchanged.

$ printf " % # x " $((0x 8765432 1 | 0x ff))0x 876543 ff