这是以利亚·卡根所支持的观点的长版本:

while [ $(( i-- )) -gt 0 ]; do echo -n "  "; done

当然，你也可以使用printf，但不是我喜欢的:

printf "%$(( i*2 ))s"

这个版本与Dash兼容:

until [ $(( i=i-1 )) -lt 0 ]; do echo -n "  "; done

I是初始数。

我怎么用echo来做呢?

如果echo后面跟着sed，你可以用echo来实现:

echo | sed -r ':a s/^(.*)$/=\1/; /^={100}$/q; ba'

实际上，这个回声在这里是不必要的。

在bash 3.0或更高版本中

for i in {1..100};do echo -n =;done

我的答案有点复杂，可能并不完美，但对于那些希望输出大数字的人来说，我能够在3秒内完成大约1000万。

repeatString(){
    # argument 1: The string to print
    # argument 2: The number of times to print
    stringToPrint=$1
    length=$2

    # Find the largest integer value of x in 2^x=(number of times to repeat) using logarithms
    power=`echo "l(${length})/l(2)" | bc -l`
    power=`echo "scale=0; ${power}/1" | bc`

    # Get the difference between the length and 2^x
    diff=`echo "${length} - 2^${power}" | bc`

    # Double the string length to the power of x
    for i in `seq "${power}"`; do 
        stringToPrint="${stringToPrint}${stringToPrint}"
    done

    #Since we know that the string is now at least bigger than half the total, grab however many more we need and add it to the string.
    stringToPrint="${stringToPrint}${stringToPrint:0:${diff}}"
    echo ${stringToPrint}
}

如果你想在echo和printf的不同实现之间遵循posix并保持一致性，和/或shell而不仅仅是bash:

seq(){ n=$1; while [ $n -le $2 ]; do echo $n; n=$((n+1)); done ;} # If you don't have it.

echo $(for each in $(seq 1 100); do printf "="; done)

.．.将在所有地方产生与perl -E 'say "=" x 100'相同的输出。

for i in {1..100}
do
  echo -n '='
done
echo