循环意外地在Bash脚本中无限运行

Question

我在我的Mac上创建了一个bash脚本Armstrong.sh.

这是一个检查数字是一个非常强的数字的函数.

# This function works properly
armstrong() {

    num=$1    # Making a copy of the received number.
    sum=0     # This will store the sum of cubes of each digit from $num



    while [ $num -gt 0];      # This loop runs while $num is greater than 0
    do
        temp=`expr $num % 10`                      # Extract the last digit of the number
        sum=`expr $sum + $temp \* $temp \* $temp`  # Cube the last digit and add it to $sum
        num=`expr $num / 10`                       # Remove the last digit of the number
    done



    if [ $sum -eq $1 ];   # If $sum == $1, i.e., If the number is armstrong
    then   
        echo "$1 is an armstrong number"        # print the number
    else
        echo "$1 is not an armstrong number"
    fi
} 

当我写下面的代码时,

armstrong 1     # this is an armstrong number
armstrong 153   # This is an armstrong number
armstrong 24    # This is not an armstrong number

然后它的输出如下,

1 is an armstrong number
153 is an armstrong number
24 is not an armstrong number

直到现在这都很好.

但问题在于此.
当我尝试使用如下循环打印范围内的所有阿姆斯特朗号码时:

# Accept start and end point of the range
echo -n "Enter start = "
read start
echo -n "Enter end = "
read end

# Loop from start to end point and call the armstrong() function
for ((num = $start; num <= $end; num++))
do
    armstrong $num    # Calling the function.
done  

所以我的问题是:
1>如何让循环按预期工作？
2>有什么办法,我可以写代码,而无需使用$temp的armstrong()功能？
就像sum += Math.pow(num%10, 3);在Java中一样？
3>请给我一个更简洁的方法来编写这个armstrong功能.

Answer 1

你的函数使用变量num而不将其声明为局部变量,因此它改变了循环引用的相同shell全局变量,从而重置了循环的状态并阻止它完成.

在功能内部,改变

num=$1

至

local num=$1

...理想情况下,对函数内的所有其他变量执行相同操作,除非您明确希望函数修改全局范围内的变量.