解釋該腳本的階段

Question

誰能解釋這個shell腳本？

#!/bin/bash
    read a
    STR=" $a   "
    ARR=($STR)
    LEN=${#ARR[*]}
    LAST=$(($LEN-1))

    ARR2=()
    I=$LAST
    while [[ $I -ge 0 ]]; do
        ARR2[ ${#ARR2[*]} ]=${ARR[$I]}   
        I=$(($I-1))
    done

    echo ${ARR2[*]}

謝謝。

Answer 1

評論到exaplain每行。

#!/bin/bash
read a # Reads from stdin
STR=" $a   " # concat the input with 1 space after and three before
ARR=($STR) # convert to array?
LEN=${#ARR[*]} # get the length of the array
LAST=$(($LEN-1)) # get the index of the last element of the array

ARR2=() # new array
I=$LAST # set var to make the first, the last 

while [[ $I -ge 0 ]]; do # reverse iteration
    ARR2[ ${#ARR2[*]} ]=${ARR[$I]}   #add the array item into new array
    I=$(($I-1)) # decrement variable
done

echo ${ARR2[*]} # echo the new array (reverse of the first)

Answer 2

格式確實弄亂了您的帖子。 這是重新格式化的腳本：

#!/bin/bash

read a
STR=" $a "
ARR=($STR)
LEN=${#ARR[*]}
LAST=$(($LEN-1))

ARR2=()
I=$LAST
while [[ $I -ge 0 ]];
    do ARR2[ ${#ARR2[*]} ]=${ARR[$I]}
    I=$(($I-1))
done

echo ${ARR2[*]}

轉過一個單詞列表會做什么。

$ echo "a b c d e f" | ./foo.sh 
f e d c b a

$ echo "The quick brown fox jumps over the lazy dog" | ./foo.sh 
dog lazy the over jumps fox brown quick The

為了描述它是如何工作的，首先找到將字符串轉換為數組，然后計算出數組中的項數，然后通過遞減I迭代遍歷該數組，然后回顯出結果數組