Why does Bash with -e option exit when 'let' expression evaluates to 0?

Question

I am struggling to understand why the bash -e option exits this script. It happens only when the expression calculated gives 0:

#!/bin/bash
set -ex
table_year=( 1979 1982 1980 1993 1995 )
year=$1 
let indice=year-1
real_year=${table_year[$indice]}
echo OK $real_year

Is is ok when:

./bash_test_array 2

but not when:

./bash_test_array 1

indice is this case equals to 0. Why the -e option causes an exit ?

https://www.gnu.org/software/bash/manual/html_node/The-Set-Builtin.html — anishsane, Mar 31 '15 at 11:17
@Joe: While the cause may be the same, the problem in this question is expressed much simpler than in that other question (assignment vs. increment side effect). — Michael Jaros, Mar 31 '15 at 11:27

Michael Jaros · Answer 1 · 2015-04-02T09:00:47.787

5

See help let:

Exit Status: If the last ARG evaluates to 0, let returns 1; let returns 0 otherwise..

The behavior of the let builtin is the same as for the commonly used expr command:

Exit status is [...] 1 if EXPRESSION is null or 0 [...]

You can use arithmetic expansion instead:

indice=$(( year - 1 ))

This statement will return 0 even if the assigned expression evaluates to 0.

edited Apr 02 '15 at 09:00

answered Mar 31 '15 at 11:16

Michael Jaros

1

Defintely better than using `let` – hek2mgl Mar 31 '15 at 11:18
1

I would add that `indice=\`expr $year - 1\`` has the same behaviour as let. – PBrockmann Apr 02 '15 at 07:32
@PBrockmann: Thanks, `expr` behavior mentioned. – Michael Jaros Apr 02 '15 at 09:01

score 1 · Answer 2 · answered Mar 31 '15 at 11:17

1

You can use the following trick:

let indice=year-1 || true

answered Mar 31 '15 at 11:17

hek2mgl

2 Answers2