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
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}"
}
其他回答
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}
}
如果你想在echo和printf的不同实现之间遵循posix并保持一致性,和/或shell而不仅仅是bash:
seq(){ n=$1; while [ $n -le $2 ]; do echo $n; n=$((n+1)); done ;} # If you don't have it.
echo $(for each in $(seq 1 100); do printf "="; done)
...将在所有地方产生与perl -E 'say "=" x 100'相同的输出。
我的建议(接受n的变量值):
n=100
seq 1 $n | xargs -I {} printf =
在bash 3.0或更高版本中
for i in {1..100};do echo -n =;done
我怎么用echo来做呢?
如果echo后面跟着sed,你可以用echo来实现:
echo | sed -r ':a s/^(.*)$/=\1/; /^={100}$/q; ba'
实际上,这个回声在这里是不必要的。