Argument string to integer in bash

In bash, one does not "convert an argument to an integer to perform arithmetic". In bash, variables are treated as integer or string depending on context.

To perform arithmetic, you should invoke the arithmetic operator $((...)). For example:

$ a=2
$ echo $a + 1
2 + 1
$ echo $(($a + 1))
3

You should be aware that the shell only performs integer arithmetic. If you have floating point numbers (numbers with decimals), then there are other tools to assist. For example, use bc:

$ b=3.14
$ echo $(($b + 1))
bash: 3.14 + 1: syntax error: invalid arithmetic operator (error token is ".14 + 1")
$ echo "$b + 1" | bc -l
4.14

In bash, you can perform the converting from anything to integer using printf -v:

printf -v int '%d\n' "$1" 2>/dev/null

Floating number will be converted to integer, while anything are not look like a number will be converted to 0. Exponentiation will be truncated to the number before e

Example:

$ printf -v int '%d\n' 123.123 2>/dev/null
$ printf '%d\n' "$int"
123
$ printf -v int '%d\n' abc 2>/dev/null
$ printf '%d\n' "$int"
0
$ printf -v int '%d\n' 1e10 2>/dev/null
$ printf '%d\n' "$int"
1