Let's say I'm writing a bash script. It will make 5 calls. Three of the calls are expected to return zero on success, one is expected to return 1 on success, and one whose return code I don't care about. Here's this script written with set -e:
#!/bin/bash set -e call0 call1 call2 call3 || ret=$? if ((ret != 1)); then exit 1 fi call4 || true Here's the same script written without the use of set -e, following the suggestion that all return codes are checked individually #!/bin/bash call0 || exit 1 call1 || exit 1 call2 || exit 1 call3 if (($? != 1)); then exit 1 fi call4 How exactly is the second script an improvement over the first? I'm seriously asking because I really don't understand. As far as I can tell, it is generally agreed upon that calls return zero on success. Using set -e is equivalent to explicitly making this single simplifying, and reasonable, assumption. Clearly there are a couple of circumstances where this is not true, but it is certainly true in the majority of cases. If it's true 90% of the time, then using set -e means we don't have to explicitly check those 90% of the calls. If we don't use set -e then we have to explicitly check those 90% of calls. AND we still have to check the return of the remaining 10% whose return we do care about. As far as I can tell, not using set -e is equivalent to writing a lot more code for no particular reason. I'm honestly curious to hear what you guys see as wrong with this reasoning.
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