bash set -e and i=0;let i++ do not agree

the answer to my question is not to use let (or shift, or...) but to use

i=$((i+1))

when trying to check a bash script by setting 'exit on non-zero status code' with

set -e

The bash manual states that set -e has the effect of 'Exit immediately if a simple command exits with a non-zero status.'.

Unfortunately let (and shift and ...) return the result of the computation ('If the last arg evaluates to 0, let returns 1; 0 is returned otherwise'). So instead of a status code one gets a return value of some sort. And sometimes this return value will be zero and sometimes one depending on the computation. Therefore set -e will cause the script to exit depending on the result of your computation!!! and there is nothing to do about it unless either you don't use it ever or resort to

let i++ || true

as pointed by arnaud576875 which btw adds extra CPU burden.

Using

let ++i

works only for the specific case that i is not -1, as with let i++ which works only for when i is not 0. Therefore half-solutions.

I love Unix though, I wouldn't have it any other way.

If the last argument of let evaluates to 0, let returns 1 (so, a non-zero status):

From the manual:

   let arg [arg ...]

Each arg is an arithmetic expression to be evaluated. If the last arg evaluates to 0, let returns 1; 0 is returned otherwise.

i++ evaluates to zero when i is 0 (because it's a post-increment, so the previous value of i is returned), so let returns 1, and due to set -e, bash exists.

Here are some solutions:

let ++i         # pre-increment, if you expect `i` to never be -1let i++ 1       # add an expression evaluating to non-zerolet i++ || true # call true if let returns non-zero

Looking at the BASH manpage on the set -e:

Exit immediately if a simple command (see SHELL GRAMMAR above) exits with a non-zero status. [...]

So, if any statement returns a non-zero exit code, the shell will exit.

Taking a look at the BASH manpage, on the let command:

If the last arg evaluates to 0, let returns 1; 0 is returned otherwise.

But wait! The answer to i++ is a one and not a zero! It should have worked!

Again, the answer is with the BASH manpage on the increment operator:

id++ id--: variable post-increment and post-decrement

Okay, not so clear. Try this shell script:

#!/bin/bashset -e -vi=1; let ++i; echo "I am still here"i=0; let ++i; echo "I am still here"i=0; ((++i)); echo "I am still here"

Hmmm... that works as expected, and all I did was change i++ to ++i in each line.

The i++ is a post-increment operator. That means, it increments i after the let statement returns a value. Since i was zero before being incremented, the let statement returns a non-zero value.

However, the ++i is a pre-increment operator. That means it increments i before returning the exit status. Since i is incremented to a 1, the exit status becomes a zero.

I hope this makes sense.