143902 2015-04-17 16:20:56 -0700 Stepping in the debugger doesn't always advance as one expects - step-over should complete the line 2026-01-12 09:08:37 -0800 1 1 1 Unclassified WebKit Web Inspector 528+ (Nightly build) Unspecified Unspecified NEW InRadar P2 Normal --- 1 dino joepeck bburg graouts inspector-bugzilla-changes joepeck jonowells webkit-bug-importer oldest_to_newest 1086472 0 251066 dino 2015-04-17 16:20:56 -0700 Created attachment 251066 Test case For single line statements, calling step in the JS debugger doesn't always advance to the next line. See the attached testcase. 1086473 1 webkit-bug-importer 2015-04-17 16:21:27 -0700 <rdar://problem/20595484> 1234532 2 joepeck 2016-09-28 13:42:13 -0700 I'll use this bug to track making "step-over" do line based stepping instead of ~statement stepping. In general, stepping should be significantly improved soon so that you will know exactly what is executing each step. There has been discussion on how step-over should behave in such cases, and it seems "skip all things on this line" is what developers are familiar with. 1275733 3 bburg 2017-02-10 20:22:42 -0800 Joe, is there any reason to keep this bug open any more? 1275770 4 joepeck 2017-02-11 01:13:56 -0800 Yes. The idea of "step-over" being line-based is still valid. Currently we can pause on multiple statements within a line. Moving to line-based (by sending the backend a start offset and end offset of the line) would be an enhancement we could add. 251066 2015-04-17 16:20:56 -0700 2015-04-17 16:20:56 -0700 Test case stepping.html text/html 344 dino PHNjcmlwdD4KCmZ1bmN0aW9uIHJ1bigpIHsKICAgIHZhciBmb28gPSAwOwogICAgY29uc29sZS5s b2coIkknbSBnb2luZyB0byBlbnRlciB0aGUgZGVidWdnZXIgaWYgaXQgaXMgb3Blbi4iKTsKICAg IGRlYnVnZ2VyOwogICAgaWYgKCFmb28pIGZvbyA9IERhdGUubm93KCk7IC8vIEhvdyBtYW55IHN0 ZXBzIHNob3VsZCBpdCB0YWtlIHRvIGdldCB0byB0aGUgbmV4dCBsaW5lPwogICAgY29uc29sZS5s b2coIkkgcmVhY2hlZCB0aGUgbmV4dCBsaW5lLiIpOwp9Cgp3aW5kb3cuYWRkRXZlbnRMaXN0ZW5l cigibG9hZCIsIHJ1biwgZmFsc2UpOwo8L3NjcmlwdD4KPHA+T3BlbiB0aGUgaW5zcGVjdG9yPC9w Pgo=