Question

此代码不会导致任何失败，但如果使用counter ++，则第一次迭代失败。

parameters="one two three"  
counter=0

  for option in $parameters
  do
    eval $option=${args[$counter]}
    ((counter = counter + 1)) # If you do ((counter++)) it fails the first iteration, weird.
    echo $option $?
  done

这就是我的意思：

ulukai@computer:~$ bash -x test.sh 
+ parameters='one two three'
+ counter=0
+ for option in '$parameters'
+ eval one=
++ one=
+ (( counter++ ))
+ echo one 1
one 1
+ for option in '$parameters'
+ eval two=
++ two=
+ (( counter++ ))
+ echo two 0
two 0
+ for option in '$parameters'
+ eval three=
++ three=
+ (( counter++ ))
+ echo three 0
three 0
ulukai@computer:~$ vi test.sh
ulukai@computer:~$ bash -x test.sh 
+ parameters='one two three'
+ counter=0
+ for option in '$parameters'
+ eval one=
++ one=
+ (( counter=counter+1 ))
+ echo one 0
one 0
+ for option in '$parameters'
+ eval two=
++ two=
+ (( counter=counter+1 ))
+ echo two 0
two 0
+ for option in '$parameters'
+ eval three=
++ three=
+ (( counter=counter+1 ))
+ echo three 0
three 0

我认为这对于理解这个问题的人来说足够解释，但是因为我需要添加更多文本才能提交这个，我正在写这一行。

Answer 1

i++将返回i的旧值，因此第一个counter++将返回0，这意味着在Bash的算术上下文中为FALSE。

（参考：https://en.wikipedia.org/wiki/Increment_and_decrement_operators）

当计数器== 0时，为什么（（计数器++））失败？

1 个答案: