CODEWITHSUNDEEP
linuxbashawk
maihabunash
maihabunash

Reputation: 1702

How to perform Arithmetic progression in bash

I need to count from number N1 until N2 with Increments of 100 For example

46500 to 49999 Increments of 100

Will print the following


46600
46700
46800
46900
47000
.
.
.
49900

Please advice how to implement (print) this counting with bash

Upvotes: 1

Views: 1491

Answers (3)

Gilles Quénot
Gilles Quénot

Reputation: 185434

Using brace expansion, a new feature of only with builtins

(for the increment part) :

printf '%s\n' {46500..49999..100}

Output

46500
46600
46700
46800
46900
47000
47100
47200
47300
(...)
49300
49400
49500
49600
49700
49800
49900

Upvotes: 4

nu11p01n73R
nu11p01n73R

Reputation: 26667

You can use seq

$ seq 46600 100 49999 
46600
46700
46800
46900
.
.
.
49600
49700
49800
49900

From man page

NAME
       seq - print a sequence of numbers

SYNOPSIS
       seq [OPTION]... LAST
       seq [OPTION]... FIRST LAST
       seq [OPTION]... FIRST INCREMENT LAST

Upvotes: 6

anubhava
anubhava

Reputation: 785376

You can use BASH's arithmetic evaluator ((...)) for this:

for ((i=46600; i<=49999; i+=100)); do echo $i; done
46500
46600
46700
...
...
49900

You can even use variable:

s=46600
e=49999
for ((i=$s; i<=$e; i+=100)); do echo $i; done

Upvotes: 3

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