Compaing integers in bash

Question

I have been using integer comparisons with standard approach

(( n == 0 ))

But this time I want to check if the variable is not zero. But I have found that != is only for strings. Is this correct? How is this integer comparison customarily approached?

Does this answer your question? [Check if bash variable equals 0](https://stackoverflow.com/questions/13086109/check-if-bash-variable-equals-0) — 0stone0, Dec 01 '21 at 10:01
It's not correct. See the bash man-page, section ARITHMETIC EVALUATION. — user1934428, Dec 01 '21 at 12:51

score 0 · Accepted Answer · answered Dec 01 '21 at 09:53

0

This works in my bash:

if (( 1 != 2 )); then echo Y; fi
Y

answered Dec 01 '21 at 09:53

D-FENS

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What about 0 != 00? That would distinguish string comparisons from numeric comparisons. – rghome Dec 01 '21 at 10:01
1

It works properly. – D-FENS Dec 01 '21 at 10:03
Are we settled on using `!=` with `(())`? Have bash superceded the previous insistence by some that `!=` is for strings. – Dilna Dec 01 '21 at 10:09
1

@Phil: As far as I can see in my test, != works with integers in (( )). Usually I use [[ ]] when comparing strings and != works there too as expected. – D-FENS Dec 01 '21 at 10:12
@rghome : 0 and 00 are numerically equal. Therefore `(( 0 != 00)) || echo false` outputs _false_. – user1934428 Dec 01 '21 at 12:54
@Phil : You do not have **any** string comparision in arithmetic context. – user1934428 Dec 01 '21 at 12:55

Compaing integers in bash

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