bash 如何在bash中做幂
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How to do exponentiation in bash
提问by user458553
I try
我试试
echo 10**2
it prints 10**2. How to make it work?
它打印10**2。如何使它工作？
回答by codaddict
You can do:
你可以做：
let var=10**2 # sets var to 100.
or even better and recommended way:
甚至更好的推荐方式：
var=$((10**2)) # sets var to 100.
If you just want to print the expression result you can do:
如果您只想打印表达式结果，您可以执行以下操作：
echo $((10**2)) # prints 100.
For large numbers you might want to use the exponentiation operator of bcas:
对于大数，您可能需要使用bcas的幂运算符：
bash:$ echo 2^100 | bc
1267650600228229401496703205376
If you want to store the above result in a variable you can again use the $(())syntax as:
如果要将上述结果存储在变量中，您可以再次使用以下$(())语法：
var=$((echo 2^100 | bc))
回答by ghostdog74
various ways
各种方式
Bash
重击
x=2
echo $((x**2))
Awk
awk
awk 'BEGIN{print 2**2}'
bc
公元前
echo "2 ^ 2" |bc
dc
直流电
dc -e '2 2 ^ p'
回答by firefly
Actually var=$((echo 2^100 | bc))doesn't work - bash is trying to do math inside (()). But a
command line sequence is there instead so it creates an error
实际上var=$((echo 2^100 | bc))不起作用 - bash 试图在里面做数学运算(())。但是有一个命令行序列，所以它会产生一个错误
var=$(echo 2^100 | bc)works as the value is the result of the command line executing inside
()
var=$(echo 2^100 | bc)值是在内部执行的命令行的结果
()
