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smart highlighting is not so smart

Open rhelmot opened this issue 4 years ago • 4 comments

image

I'm guessing it doesn't highlight the second variable usage because it views the assignment as killing the previous value, but the first one is an access by reference, so it should apply to all usages of the variable in this function, regardless of where it gets killed.

rhelmot avatar Dec 19 '20 20:12 rhelmot

s_10 at 0x48cb2 is a new def regardless of how s_10 was used previously. At 0x48ca9, s_10 is a use of the previous def.

You may switch to less-smart highlighting by pressing a.

ltfish avatar Dec 20 '20 00:12 ltfish

right... but 0x48ca9 isn't strictly a use of the variable. it's a reference which may be used as many times in the future as it cares to be, including after the current def is killed. Intuitively, I would like to see that reflected in the highlighting

rhelmot avatar Dec 20 '20 02:12 rhelmot

This issue has been marked as stale because it has no recent activity. Please comment or add the pinned tag to prevent this issue from being closed.

github-actions[bot] avatar Jun 21 '22 02:06 github-actions[bot]

This issue has been marked as stale because it has no recent activity. Please comment or add the pinned tag to prevent this issue from being closed.

github-actions[bot] avatar Oct 20 '22 02:10 github-actions[bot]