bash 计算命令行参数的总和并显示结果
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Calculate sum of command line arguments and display the result
提问by classA
Write a bash script that adds its command line arguments and display the result. Assume command line arguments are all integers apart from the argument itself.
编写一个 bash 脚本，添加其命令行参数并显示结果。假设命令行参数除了参数本身之外都是整数。
So for example:
例如：
% addprogram 2 5 7
The total is 14
I've tried all I can and can't seem to find anything on the net, please keep the answer really simple.
我已经尽我所能，但似乎在网上找不到任何东西，请保持答案非常简单。
I've tried stuff like:
我试过这样的东西：
#!/bin/bash
total += $@
But I don't think that operator exists in bash.
但我不认为该运算符存在于 bash 中。
My algorithm would be something like this:
我的算法是这样的：
- For argument passed from the command line,
- Add the first argument to the total,
- move onto the next argument and add that to the total,
- repeat as long as there are arguments
- output total
- 对于从命令行传递的参数，
- 将第一个参数添加到总数中，
- 转到下一个参数并将其添加到总数中，
- 只要有参数就重复
- 总产量
回答by
Because nobody's done it yet
因为还没有人做过
IFS=+
bc<<<"$*"
Using bcgets you bignums; shell builtin arithmetic fails when the numbers are huge.
使用bc让你大数；当数字很大时，shell 内置算术失败。
exprmay or may not do bignums depending on whether it's a recent GNU coreutils version.
expr可能会或可能不会做 bignums，这取决于它是否是最近的 GNU coreutils 版本。
The IFSand <<<are just tricks to make it shorter. A more understandable version would be
该IFS和<<<只是技巧，使其更短。一个更容易理解的版本是
echo $* | sed 's/ /+/g' | bc
回答by Micha? ?rajer
posix compatible version:
posix 兼容版本：
for i do
sum=$(expr $sum + $i)
done
echo $sum
Note that by default foriterates over $@.
请注意，默认情况下for迭代$@.
回答by mogul
Something like this?
像这样的东西？
for n in $@
do
let sum=$sum+$n
done
echo The total is $sum
回答by Micha? Kosmulski
I like awk, so I'd do it like this:
我喜欢awk，所以我会这样做：
echo $@ | tr ' ' '\n' | awk '{sum+=echo $@ | tr ' ' '\n' | ( sum=0; while read; do sum=$((sum+REPLY)); done; echo $sum )
}END{print sum}'
And a pure-bash version could go like:
一个纯 bash 版本可能是这样的：
sum=0
while [ -n "" ]; do
sum=`expr $sum + `
shift
done
echo $sum
回答by Adam Siemion
Following the steps you have provided, with shiftand expr:
按照您提供的步骤，使用shift和expr：
total=0
for i in $@; do
(( total+=i ))
done
echo "The total is $total"
回答by Ansgar Wiechers
I'd do it like this:
我会这样做：
##代码##