for循环在bash中退出到早期，没有错误。

Question

为什么我的for循环: for ((i=1，;i<=$n，i++))不能工作？我搞不懂这事。例如，当n为4时，行echo $n返回4，但不会再次进入循环。我也没有任何错误。我试着做一个小循环，比如：

for ((i=1, i<=$n; i++)); do 
    echo "this works" 
done

这件事做得很好，对我来说就更奇怪了。提前感谢

read n
length=16
p=()
p[1]=50
rest=63
function s() {
    arr=($@)
    line="_____________________________________________________________________"
    for i in ${arr[@]}; do
        line=$( echo $line |  sed s/./1/$i)
    done
    echo $line
}
for ((i=1; i<=$n; i++)); do
    echo $n
    for ((j=1; j<=$length; j++)); do
        s ${p[@]}
    done
    len=${#p[@]}
    ((len=$len*2))
    for ((k=1; k<=$len; k+=2)); do
        ((p[$k+1]=p[$k]+1))
        ((p[$k]=p[$k]-1))
    done
    for ((l=1; l<=$length; l++)); do
        s ${p[@]}
        len=${#p[@]}
        for ((m=1; m<=$len; m+=2)); do
            ((p[$m+1]=p[$m+1]+1))
            ((p[$m]=p[$m]-1))
        done
    done
    ((rest=$rest-2*$length))
    ((length=$length/2))
done

Answer 1

如果不声明变量为局部变量，它们是全局的--因此s函数中的循环覆盖函数外部使用的相同的i计数器，从而导致外部循环的早期退出。

考虑使用以下代码：

s() {
    local -a arr  # declare a function-local array (sparse, like all bash arrays)
    local i       # ...and a function-local counter

    for i; do     # by default, a for statement iterates over "$@"; that works for us.
      arr[$i]=1   # specifically set a value only for named items
    done

    # ...thereafter, iterate through the range of characters we want to print...
    for ((i=0; i<70; i++)); do
      printf '%s' "${arr[$i]:-_}" # and print the array entry if present, or an _ otherwise
    done
    printf '\n'                   # ...followed by a trailing newline.
}