How to convert string to integer in UNIX

Question

I have d1=\"11\" and d2=\"07\". I want to convert d1 and d2 to integers and perform d1-d2. How do I do this

Answer 1

An answer that is not limited to the OP's case

The title of the question leads people here, so I decided to answer that question for everyone else since the OP's described case was so limited.

TL;DR

I finally settled on writing a function.

1. If you want 0 in case of non-int:

int(){ printf '%d' ${1:-} 2>/dev/null || :; }

2. If you want [empty_string] in case of non-int:

int(){ expr 0 + ${1:-} 2>/dev/null||:; }

3. If you want find the first int or [empty_string]:

int(){ expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null||:; }

4. If you want find the first int or 0:

# This is a combination of numbers 1 and 2
int(){ expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null||:; }

If you want to get a non-zero status code on non-int, remove the ||: (aka or true) but leave the ;

Tests

# Wrapped in parens to call a subprocess and not `set` options in the main bash process
# In other words, you can literally copy-paste this code block into your shell to test
( set -eu;
    tests=( 4 "5" "6foo" "bar7" "foo8.9bar" "baz" " " "" )
    test(){ echo; type int; for test in "${tests[@]}"; do echo "got '$(int $test)' from '$test'"; done; echo "got '$(int)' with no argument"; }

    int(){ printf '%d' ${1:-} 2>/dev/null||:; };
    test

    int(){ expr 0 + ${1:-} 2>/dev/null||:; }
    test

    int(){ expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null||:; }
    test

    int(){ printf '%d' $(expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null)||:; }
    test

    # unexpected inconsistent results from `bc`
    int(){ bc<<<"${1:-}" 2>/dev/null||:; }
    test
)

Test output

int is a function
int ()
{
    printf '%d' ${1:-} 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '0' from '6foo'
got '0' from 'bar7'
got '0' from 'foo8.9bar'
got '0' from 'baz'
got '0' from ' '
got '0' from ''
got '0' with no argument

int is a function
int ()
{
    expr 0 + ${1:-} 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '' from '6foo'
got '' from 'bar7'
got '' from 'foo8.9bar'
got '' from 'baz'
got '' from ' '
got '' from ''
got '' with no argument

int is a function
int ()
{
    expr ${1:-} : '[^0-9]*\([0-9]*\)' 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '6' from '6foo'
got '7' from 'bar7'
got '8' from 'foo8.9bar'
got '' from 'baz'
got '' from ' '
got '' from ''
got '' with no argument

int is a function
int ()
{
    printf '%d' $(expr ${1:-} : '[^0-9]*\([0-9]*\)' 2>/dev/null) || :
}
got '4' from '4'
got '5' from '5'
got '6' from '6foo'
got '7' from 'bar7'
got '8' from 'foo8.9bar'
got '0' from 'baz'
got '0' from ' '
got '0' from ''
got '0' with no argument

int is a function
int ()
{
    bc <<< "${1:-}" 2> /dev/null || :
}
got '4' from '4'
got '5' from '5'
got '' from '6foo'
got '0' from 'bar7'
got '' from 'foo8.9bar'
got '0' from 'baz'
got '' from ' '
got '' from ''
got '' with no argument

Note

I got sent down this rabbit hole because the accepted answer is not compatible with set -o nounset (aka set -u)

# This works
$ ( number="3"; string="foo"; echo $((number)) $((string)); )
3 0

# This doesn't
$ ( set -u; number="3"; string="foo"; echo $((number)) $((string)); )
-bash: foo: unbound variable

Answer 2

Use this:

#include <stdlib.h>
#include <string.h>

int main()
{
    const char *d1 = "11";
    int d1int = atoi(d1);
    printf("d1 = %d\n", d1);
    return 0;
}

etc.

Answer 3

let d=d1-d2;echo $d;

This should help.

Answer 4

The standard solution:

 expr $d1 - $d2

You can also do:

echo $(( d1 - d2 ))

but beware that this will treat 07 as an octal number! (so 07 is the same as 7, but 010 is different than 10).

Answer 5

Any of these will work from the shell command line. bc is probably your most straight forward solution though.

Using bc:

$ echo "$d1 - $d2" | bc

Using awk:

$ echo $d1 $d2 | awk '{print $1 - $2}'

Using perl:

$ perl -E "say $d1 - $d2"

Using Python:

$ python -c "print $d1 - $d2"

all return

4