如何在 bash 中对十六进制数执行按位运算？
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How to perform bitwise operations on hexadecimal numbers in bash?
提问by Jimmy Xu
In my bashscript I have a string containing a hexadecimal number, e.g. hex="0x12345678". Is it possible to treat it as a hex number and do bit shifting on it?
在我的bash脚本中，我有一个包含十六进制数的字符串，例如hex="0x12345678". 是否可以将其视为十六进制数并对其进行位移？
采纳答案by that other guy
You can easily bitshift such numbers in an arithmetic context:
您可以在算术上下文中轻松地对此类数字进行位移：
$ hex="0x12345678"
$ result=$((hex << 1))
$ printf "Result in hex notation: 0x%x\n" "$result"
0x2468acf0
回答by Isaac
Of course you can do bitwise operations (inside an Arithmetic Expansion):
当然，您可以进行按位运算（在算术扩展中）：
$ echo "$((0x12345678 << 1))"
610839792
Or:
或者：
$ echo "$(( 16#12345678 << 1 ))"
610839792
The value could be set in a variable as well:
该值也可以设置在变量中：
$ var=0x12345678 # or var=16#12345678
$ echo "$(( var << 1 ))"
610839792
And you can do OR, AND and XOR:
你可以做 OR、AND 和 XOR：
$ echo "$(( 0x123456 | 0x876543 ))"
9925975
And to get the result in hex as well:
并以十六进制形式获得结果：
$ printf '%X\n' "$(( 0x12345678 | 0xDEADBEEF ))" # Bitwise OR
DEBDFEFF
$ printf '%X\n' "$(( 0x12345678 & 0xDEADBEEF ))" # Bitwise AND
12241668
$ printf '%X\n' "$(( 0x12345678 ^ 0xDEADBEEF ))" # Bitwise XOR
CC99E897
