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}
}

一种纯粹的Bash方式，没有eval，没有subshell，没有外部工具，没有大括号展开(即，你可以在变量中重复数字):

如果给你一个变量n，它展开为一个(非负的)数字和一个变量模式，例如，

$ n=5
$ pattern=hello
$ printf -v output '%*s' "$n"
$ output=${output// /$pattern}
$ echo "$output"
hellohellohellohellohello

你可以用它来创建一个函数:

repeat() {
    # $1=number of patterns to repeat
    # $2=pattern
    # $3=output variable name
    local tmp
    printf -v tmp '%*s' "$1"
    printf -v "$3" '%s' "${tmp// /$2}"
}

这套:

$ repeat 5 hello output
$ echo "$output"
hellohellohellohellohello

对于这个小技巧，我们经常使用printf:

-v varname: instead of printing to standard output, printf will put the content of the formatted string in variable varname. '%*s': printf will use the argument to print the corresponding number of spaces. E.g., printf '%*s' 42 will print 42 spaces. Finally, when we have the wanted number of spaces in our variable, we use a parameter expansion to replace all the spaces by our pattern: ${var// /$pattern} will expand to the expansion of var with all the spaces replaced by the expansion of $pattern.

你也可以通过间接展开来去掉repeat函数中的tmp变量:

repeat() {
    # $1=number of patterns to repeat
    # $2=pattern
    # $3=output variable name
    printf -v "$3" '%*s' "$1"
    printf -v "$3" '%s' "${!3// /$2}"
}

如果你想重复一个字符n次，n是一个变量的次数，这取决于，比如说，字符串的长度，你可以这样做:

#!/bin/bash
vari='AB'
n=$(expr 10 - length $vari)
echo 'vari equals.............................: '$vari
echo 'Up to 10 positions I must fill with.....: '$n' equal signs'
echo $vari$(perl -E 'say "=" x '$n)

它显示:

vari equals.............................: AB  
Up to 10 positions I must fill with.....: 8 equal signs  
AB========  

最简单的方法是在csh/tcsh中使用这一行代码:

printf "%50s\n" '' | tr '[:blank:]' '[=]'

另一种选择是使用GNU seq并删除它生成的所有数字和换行:

seq -f'#%.0f' 100 | tr -d '\n0123456789'

这个命令打印#字符100次。

稍微长一点的版本，但如果你出于某种原因必须使用纯Bash，你可以使用一个带增量变量的while循环:

n=0; while [ $n -lt 100 ]; do n=$((n+1)); echo -n '='; done