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

不是堆砌，而是另一种纯bash方法利用了数组的${//}替换:

$ arr=({1..100})
$ printf '%s' "${arr[@]/*/=}"
====================================================================================================

没有简单的方法。但是举个例子:

seq -s= 100|tr -d '[:digit:]'
# Editor's note: This requires BSD seq, and breaks with GNU seq (see comments)

或者是一种符合标准的方式:

printf %100s |tr " " "="

还有一个tput代表，但对于我手头的终端(xterm和linux)，它们似乎不支持它:)

我猜这个问题的最初目的是仅用shell的内置命令来完成这个任务。因此，for循环和printfs是合法的，而下面的rep、perl和jot则不是。但是，下面的命令

$((COLUMNS/2))

例如,打印的窗户完全线 \/\/\/\/\/\/\/\/\/\/\/\/

向@gniourf_gniourf致敬。

注意:这个答案并没有回答最初的问题，而是通过比较性能来补充现有的有用答案。

解决方案只在执行速度方面进行比较-内存需求没有考虑在内(它们在不同的解决方案中有所不同，并且可能与大量重复计数有关)。

简介:

If your repeat count is small, say up to around 100, it's worth going with the Bash-only solutions, as the startup cost of external utilities matters, especially Perl's. Pragmatically speaking, however, if you only need one instance of repeating characters, all existing solutions may be fine. With large repeat counts, use external utilities, as they'll be much faster. In particular, avoid Bash's global substring replacement with large strings (e.g., ${var// /=}), as it is prohibitively slow.

以下是在一台配有3.2 GHz Intel酷睿i5 CPU和Fusion Drive的2012年末iMac上进行的计时，运行OSX 10.10.4和bash 3.2.57，是1000次运行的平均值。

条目如下:

按执行时间升序列出(最快的先) 前缀: 米……一个潜在的多字符解决方案 年代……单字符解决方案 P…posix兼容的解决方案 接下来是对解决方案的简要描述 以原始答案的作者的名字作为后缀

小重复计数:100

[M, P] printf %.s= [dogbane]:                           0.0002
[M   ] printf + bash global substr. replacement [Tim]:  0.0005
[M   ] echo -n - brace expansion loop [eugene y]:       0.0007
[M   ] echo -n - arithmetic loop [Eliah Kagan]:         0.0013
[M   ] seq -f [Sam Salisbury]:                          0.0016
[M   ] jot -b [Stefan Ludwig]:                          0.0016
[M   ] awk - $(count+1)="=" [Steven Penny (variant)]:   0.0019
[M, P] awk - while loop [Steven Penny]:                 0.0019
[S   ] printf + tr [user332325]:                        0.0021
[S   ] head + tr [eugene y]:                            0.0021
[S, P] dd + tr [mklement0]:                             0.0021
[M   ] printf + sed [user332325 (comment)]:             0.0021
[M   ] mawk - $(count+1)="=" [Steven Penny (variant)]:  0.0025
[M, P] mawk - while loop [Steven Penny]:                0.0026
[M   ] gawk - $(count+1)="=" [Steven Penny (variant)]:  0.0028
[M, P] gawk - while loop [Steven Penny]:                0.0028
[M   ] yes + head + tr [Digital Trauma]:                0.0029
[M   ] Perl [sid_com]:                                  0.0059

只有bash的解决方案领先的包-但只有重复计数这么小!(见下文)。 外部实用程序的启动成本在这里很重要，特别是Perl的启动成本。如果必须在循环中调用它——每次迭代中重复次数很少——请避免使用multi-utility、awk和perl解决方案。

大重复计数:1000000(100万)

[M   ] Perl [sid_com]:                                  0.0067
[M   ] mawk - $(count+1)="=" [Steven Penny (variant)]:  0.0254
[M   ] gawk - $(count+1)="=" [Steven Penny (variant)]:  0.0599
[S   ] head + tr [eugene y]:                            0.1143
[S, P] dd + tr [mklement0]:                             0.1144
[S   ] printf + tr [user332325]:                        0.1164
[M, P] mawk - while loop [Steven Penny]:                0.1434
[M   ] seq -f [Sam Salisbury]:                          0.1452
[M   ] jot -b [Stefan Ludwig]:                          0.1690
[M   ] printf + sed [user332325 (comment)]:             0.1735
[M   ] yes + head + tr [Digital Trauma]:                0.1883
[M, P] gawk - while loop [Steven Penny]:                0.2493
[M   ] awk - $(count+1)="=" [Steven Penny (variant)]:   0.2614
[M, P] awk - while loop [Steven Penny]:                 0.3211
[M, P] printf %.s= [dogbane]:                           2.4565
[M   ] echo -n - brace expansion loop [eugene y]:       7.5877
[M   ] echo -n - arithmetic loop [Eliah Kagan]:         13.5426
[M   ] printf + bash global substr. replacement [Tim]:  n/a

The Perl solution from the question is by far the fastest. Bash's global string-replacement (${foo// /=}) is inexplicably excruciatingly slow with large strings, and has been taken out of the running (took around 50 minutes(!) in Bash 4.3.30, and even longer in Bash 3.2.57 - I never waited for it to finish). Bash loops are slow, and arithmetic loops ((( i= 0; ... ))) are slower than brace-expanded ones ({1..n}) - though arithmetic loops are more memory-efficient. awk refers to BSD awk (as also found on OSX) - it's noticeably slower than gawk (GNU Awk) and especially mawk. Note that with large counts and multi-char. strings, memory consumption can become a consideration - the approaches differ in that respect.

下面是生成上述代码的Bash脚本(testrepeat)。 它有两个参数:

字符重复计数 可选地，要执行的测试运行的数量，并从中计算平均时间

换句话说:上面的时间是用testrepeat 100 1000和testrepeat 1000000 1000得到的

#!/usr/bin/env bash

title() { printf '%s:\t' "$1"; }

TIMEFORMAT=$'%6Rs'

# The number of repetitions of the input chars. to produce
COUNT_REPETITIONS=${1?Arguments: <charRepeatCount> [<testRunCount>]}

# The number of test runs to perform to derive the average timing from.
COUNT_RUNS=${2:-1}

# Discard the (stdout) output generated by default.
# If you want to check the results, replace '/dev/null' on the following
# line with a prefix path to which a running index starting with 1 will
# be appended for each test run; e.g., outFilePrefix='outfile', which
# will produce outfile1, outfile2, ...
outFilePrefix=/dev/null

{

  outFile=$outFilePrefix
  ndx=0

  title '[M, P] printf %.s= [dogbane]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  # !! In order to use brace expansion with a variable, we must use `eval`.
  eval "
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    printf '%.s=' {1..$COUNT_REPETITIONS} >"$outFile"
  done"

  title '[M   ] echo -n - arithmetic loop [Eliah Kagan]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    for ((i=0; i<COUNT_REPETITIONS; ++i)); do echo -n =; done >"$outFile"
  done


  title '[M   ] echo -n - brace expansion loop [eugene y]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  # !! In order to use brace expansion with a variable, we must use `eval`.
  eval "
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    for i in {1..$COUNT_REPETITIONS}; do echo -n =; done >"$outFile"
  done
  "

  title '[M   ] printf + sed [user332325 (comment)]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    printf "%${COUNT_REPETITIONS}s" | sed 's/ /=/g' >"$outFile"
  done


  title '[S   ] printf + tr [user332325]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    printf "%${COUNT_REPETITIONS}s" | tr ' ' '='  >"$outFile"
  done


  title '[S   ] head + tr [eugene y]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    head -c $COUNT_REPETITIONS < /dev/zero | tr '\0' '=' >"$outFile"
  done


  title '[M   ] seq -f [Sam Salisbury]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    seq -f '=' -s '' $COUNT_REPETITIONS >"$outFile"
  done


  title '[M   ] jot -b [Stefan Ludwig]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    jot -s '' -b '=' $COUNT_REPETITIONS >"$outFile"
  done


  title '[M   ] yes + head + tr [Digital Trauma]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    yes = | head -$COUNT_REPETITIONS | tr -d '\n'  >"$outFile"
  done

  title '[M   ] Perl [sid_com]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    perl -e "print \"=\" x $COUNT_REPETITIONS" >"$outFile"
  done

  title '[S, P] dd + tr [mklement0]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  time for (( n = 0; n < COUNT_RUNS; n++ )); do 
    dd if=/dev/zero bs=$COUNT_REPETITIONS count=1 2>/dev/null | tr '\0' "=" >"$outFile"
  done

  # !! On OSX, awk is BSD awk, and mawk and gawk were installed later.
  # !! On Linux systems, awk may refer to either mawk or gawk.
  for awkBin in awk mawk gawk; do
    if [[ -x $(command -v $awkBin) ]]; then

      title "[M   ] $awkBin"' - $(count+1)="=" [Steven Penny (variant)]'
      [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
      time for (( n = 0; n < COUNT_RUNS; n++ )); do 
        $awkBin -v count=$COUNT_REPETITIONS 'BEGIN { OFS="="; $(count+1)=""; print }' >"$outFile"
      done

      title "[M, P] $awkBin"' - while loop [Steven Penny]'
      [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
      time for (( n = 0; n < COUNT_RUNS; n++ )); do 
        $awkBin -v count=$COUNT_REPETITIONS 'BEGIN { while (i++ < count) printf "=" }' >"$outFile"
      done

    fi
  done

  title '[M   ] printf + bash global substr. replacement [Tim]'
  [[ $outFile != '/dev/null' ]] && outFile="$outFilePrefix$((++ndx))"
  # !! In Bash 4.3.30 a single run with repeat count of 1 million took almost
  # !! 50 *minutes*(!) to complete; n Bash 3.2.57 it's seemingly even slower -
  # !! didn't wait for it to finish.
  # !! Thus, this test is skipped for counts that are likely to be much slower
  # !! than the other tests.
  skip=0
  [[ $BASH_VERSINFO -le 3 && COUNT_REPETITIONS -gt 1000 ]] && skip=1
  [[ $BASH_VERSINFO -eq 4 && COUNT_REPETITIONS -gt 10000 ]] && skip=1
  if (( skip )); then
    echo 'n/a' >&2
  else
    time for (( n = 0; n < COUNT_RUNS; n++ )); do 
      { printf -v t "%${COUNT_REPETITIONS}s" '='; printf %s "${t// /=}"; } >"$outFile"
    done
  fi
} 2>&1 | 
 sort -t$'\t' -k2,2n | 
   awk -F $'\t' -v count=$COUNT_RUNS '{ 
    printf "%s\t", $1; 
    if ($2 ~ "^n/a") { print $2 } else { printf "%.4f\n", $2 / count }}' |
     column -s$'\t' -t

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。