bash 如何在shell中将十六进制数与十六进制数进行比较？
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How to compare hexadecimal numbers with hexadecimal numbers in shell?
提问by bd1257
How to compare hexadecimal number with hexadecimal numbers in shell?
如何将十六进制数与shell中的十六进制数进行比较？
采纳答案by thkala
At least bashsupports hexadecimal integers directly, provided that they are prefixed with 0x:
至少bash直接支持十六进制整数，前提是它们带有前缀0x：
$ [[ 0xdead -lt 0xcafe ]] && echo yes || echo no
no
$ [[ 0xdead -gt 0xcafe ]] && echo yes || echo no
yes
You just use the comparison operators normally...
您通常只需使用比较运算符...
回答by neoben
You can convert your hex numbers into decimals using printfand then you can compare the numeric values, e.g.:
您可以使用将十六进制数字转换为小数printf，然后您可以比较数值，例如：
x="0xdead"
y="0xcafe"
x_num=$(printf "%d" "$x")
y_num=$(printf "%d" "$y")
if [ $x_num -gt $y_num ]; then
echo "x is my value"
else
echo "x is not my value"
fi
回答by cnicutar
How about
怎么样
(( "$answer" == 0x42 ))
echo $?
answer=0xDEADCAFE
(( "$answer" == 0xDEADCAFE ))
echo $?
