fix precedence issue in uniqnum.t #134
Merged
Conversation
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
The test `!$nanish != $NaN` doesn't make much sense: It checks whether a boolean (effectively 0 or 1) is not equal to `$NaN`, which is always true. (Also, we know `$nanish != 0` from the preceding test, so `!$nanish` could be hardcoded as 0 here.) Instead test whether `$NaN` correctly round-trips through strings, i.e. whether `$nanish` behaves like a proper NaN again (being non-equal to both the original `$NaN` and itself).
|
The Perl 5.6 CI machine got upset, but it looks like a platform setup sort of upset and not related to this change. I'll ignore that one. |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Add this suggestion to a batch that can be applied as a single commit.
This suggestion is invalid because no changes were made to the code.
Suggestions cannot be applied while the pull request is closed.
Suggestions cannot be applied while viewing a subset of changes.
Only one suggestion per line can be applied in a batch.
Add this suggestion to a batch that can be applied as a single commit.
Applying suggestions on deleted lines is not supported.
You must change the existing code in this line in order to create a valid suggestion.
Outdated suggestions cannot be applied.
This suggestion has been applied or marked resolved.
Suggestions cannot be applied from pending reviews.
Suggestions cannot be applied on multi-line comments.
Suggestions cannot be applied while the pull request is queued to merge.
Suggestion cannot be applied right now. Please check back later.
The test
!$nanish != $NaNdoesn't make much sense: It checks whether a boolean (effectively 0 or 1) is not equal to$NaN, which is always true. (Also, we know$nanish != 0from the preceding test, so!$nanishcould be hardcoded as 0 here.)Instead test whether
$NaNcorrectly round-trips through strings, i.e. whether$nanishbehaves like a proper NaN again (being non-equal to both the original$NaNand itself).