10 нояб. 2012 г. · Brace expansion occurs before variables are expanded. In order to accomplish your goal, you need to use C-style for loop.

27 сент. 2015 г. · In bash, you can perform the converting from anything to integer using printf -v: printf -v int '%d\n' "$1" 2>/dev/null

23 апр. 2016 г. · There are two ways to define and use variables of integer type in Bash. My questions: What are the differences between the variables created by the two ways?

27 сент. 2022 г. · I have a set of piped commands that generates an integer expression. A sample looks like (1 +(0x1f+0x02)) I can evaluate this expression (get the resulting ...

6 янв. 2020 г. · I'm trying to create a new variable where numbers between 0 and 9 should always be represented as 00...09 instead of 0...9.

16 авг. 2018 г. · I want to sed using a variable. For example, sed -n '1,10p' file. So, instead of 10, I want to use $k where k is an integer.

13 июл. 2012 г. · My understanding is that variables are untyped in Bash, and that any variable may be used as an integer if it only contains digits. But I get the error " ...

14 июн. 2012 г. · Using printf ability to print floats we can extend most shells to do floating point math albeit with a limited range (no more than 10 digits).

17 мар. 2020 г. · I'm trying to create filenames with a bash script, based on title, year, season and episode number. Only the title can be assured to be always present.

22 янв. 2018 г. · You can do this two ways: With ksh93-compatible shells (ksh93, zsh, bash): for (( i=1;i<=$2;i++ )) do echo "Welcome $i times" done.