Destroyed SC2181 tell inverting return value will overwrite it (markdown)

Vidar Holen
2018-10-21 17:54:09 -07:00
parent 9bb672cff2
commit 5a9a20478f

@@ -1,44 +0,0 @@
## Check exit code directly with e.g. 'if mycmd;', not indirectly with $?.
### Problematic code:
```sh
make mytarget
if [ $? -ne 0 ]
then
echo "Build failed"
fi
```
### Correct code:
```sh
if ! make mytarget
then
echo "Build failed"
fi
```
### Rationale:
Running a command and then checking its exit status `$?` against 0 is redundant.
Instead of just checking the exit code of a command, it checks the exit code of a command (e.g. `[`) that checks the exit code of a command.
Apart from the redundancy, there are other reasons to avoid this pattern:
* Since the command and its status test are decoupled, inserting an innocent command like `echo "make finished"` after `make` will cause the `if` statement to silently start comparing `echo`'s status instead.
* Scripts that run or are called with `set -e ` aka `errexit` will exit immediately if the command fails, even though they're followed by a clause that handles failure.
* The value of `$?` is overwritten by `[`/`[[`, so you can't get the original value in the relevant then/else block (e.g. `if mycmd; then echo "Success"; else echo "Failed with $?"; fi`).
To check that a command returns success, use `if mycommand; then ...`.
To check that a command returns failure, use `if ! mycommand; then ...`. Notice that `!` will overwrite `$?` value.
To additionally capture output with command substitution: `if output=$(mycommand); then ...`
This also applies to `while`/`until` loops.
### Exceptions:
The default Solaris 10 bourne shell does not support '!' outside of the test command (`if ! mycommand; then ...` returns `!: not found`)