A strange error of 'let' in bash

Question

I wonder if the code below has wrong syntax ?

#!/bin/bash
set -e
let "time_used = 1 - 1"
echo $time_used

When I run it, nothing print. The script died on let "time_used = 1 - 1".

If I remove set -e in second line, I get the correct result 0.

Why it happened?

You could figure out by trying for non-zero cases, like `let "time_used = 2 - 1"` in which case you would see `echo` output — Inian, Dec 27 '17 at 09:42

Nahuel Fouilleul · Accepted Answer · 2017-12-27T09:53:07.670

8

let is a bash built-in for shell arithmentic

let "time_used = 1 - 1"

is equivalent to

(( time_used = 1 - 1 ))

however 0 in shell arithmetic means false and gives error exit status to avoid to exit with -e || true can be added after command

(( 0 )) || true
let "time_used = 1 - 1" || true

|| true allows to "bypass" -e option for commands returning an error exit status however we can't distinguish a command that failed from a command that returns a false exit status. Other option can be to use arithmetic to return always a truthy value.

(( (time_used = 1 - 1) || 1))

edited Dec 27 '17 at 09:53

answered Dec 27 '17 at 09:43

Nahuel Fouilleul

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OP is using `set -e`. – PesaThe Dec 27 '17 at 09:45
It does not! see the other answer – Inian Dec 27 '17 at 09:45
ok, i was updating my answer – Nahuel Fouilleul Dec 27 '17 at 09:46
Thanks very much. Why is it designed to be like this? Is this a bad design? – jinge Dec 27 '17 at 10:10
`-e` option is not the default, it seems more tedious but it is better to explicitely exit on error for each command: `command to check || exit 1` can be sufficient because generally a failing command writes message on stderr otherwise to add a specific message defining a function `command || exit_fail "message" ` – Nahuel Fouilleul Dec 27 '17 at 10:14
it is the design because an arithmetic expression can be used in a for or while loop or in if then elif else statement,also it is not specific to bash but for all posix shells – Nahuel Fouilleul Dec 27 '17 at 10:18

score 3 · Answer 2 · answered Dec 27 '17 at 09:41

3

Because let returns 1 if it's result is equal 0. In other case, it returns 0.

$ help let | tail -n2 Exit Status: If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise.

answered Dec 27 '17 at 09:41

Viktor Khilin

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A strange error of 'let' in bash

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