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
for i in {1..100}
do
echo -n '='
done
echo
其他回答
另一个使用printf和tr的bash解决方案
nb。在开始之前:
我们需要另一个答案吗?可能不会。 答案已经在这里了吗?看不见,就这样。
使用printf的前导零填充特性,并使用tr转换零。这避免了任何{1..N}发电机:
$ printf '%040s' | tr '0' '='
========================================
设置宽度为'N'字符,并自定义打印的字符:
#!/usr/bin/env bash
N=40
C='-'
printf "%0${N}s" | tr '0' "${C}"
对于大N,这比生成器的性能要好得多;在我的机器上(bash 3.2.57):
$ time printf '=%.0s' {1..1000000} real: 0m2.580s
$ time printf '%01000000s' | tr '0' '=' real: 0m0.577s
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========
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
在bash 3.0或更高版本中
for i in {1..100};do echo -n =;done