我怎么用echo来做呢?

perl -E 'say "=" x 100'

当前回答

repeat() {
    # $1=number of patterns to repeat
    # $2=pattern
    printf -v "TEMP" '%*s' "$1"
    echo ${TEMP// /$2}
}

其他回答

在bash 3.0或更高版本中

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

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。

repeat() {
    # $1=number of patterns to repeat
    # $2=pattern
    printf -v "TEMP" '%*s' "$1"
    echo ${TEMP// /$2}
}
#!/usr/bin/awk -f
BEGIN {
  OFS = "="
  NF = 100
  print
}

Or

#!/usr/bin/awk -f
BEGIN {
  while (z++ < 100) printf "="
}

例子

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}"
}