我怎么用echo来做呢?

perl -E 'say "=" x 100'

当前回答

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}
}

其他回答

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.

最简单的方法是在bash中使用这一行代码:

seq 10 | xargs -n 1 | xargs -I {} echo -n  ===\>;echo

最简单的方法是在csh/tcsh中使用这一行代码:

printf "%50s\n" '' | tr '[:blank:]' '[=]'

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。

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:)