我怎么用echo来做呢?
perl -E 'say "=" x 100'
我怎么用echo来做呢?
perl -E 'say "=" x 100'
当前回答
我的建议(接受n的变量值):
n=100
seq 1 $n | xargs -I {} printf =
其他回答
不是堆砌,而是另一种纯bash方法利用了数组的${//}替换:
$ arr=({1..100})
$ printf '%s' "${arr[@]/*/=}"
====================================================================================================
没有简单的办法。使用printf和substitution避免循环。
str=$(printf "%40s")
echo ${str// /rep}
# echoes "rep" 40 times.
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}"
}
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:)
repeat() {
# $1=number of patterns to repeat
# $2=pattern
printf -v "TEMP" '%*s' "$1"
echo ${TEMP// /$2}
}