Fix a bug when multiple interfaces were parsed #104
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Fix petergtz#103 The loop used to update an out-scopped variable on it's first iteration, and all the subsequent iteration failed to go into `if structName == "" {`. The result was all the generated interface had the same name (even if the implementation was distinct for each) The loop was writing to a variable out of the scope and the second time it goes through `if structName == "" {` the condition was never false because the out of scope variable was updated with the first
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Hey @anthonyraymond, super cool that you've already fixed the issue you opened. I'm happy to merge it. Before I do that, what do you think, would you be able to also add a test for it? It's probably not super trivial, because all the tests so far simply use the one interface "Display" to do their testing. Adding another one might require a few more changes. It looks to me like a new test could either go into Curious to get your feedback. Thanks, |
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HI @petergtz, thanks for your answer. I was going to add a test in the first place, but i don't understood how to compile the project in the first place, so i had to break it appart to kind of adapt it the "module way", that was obviously not the good way to make this project module compliant ^^ But if you plan to officially migrate to module i'd be please to add a unit test for this |
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Hey @anthonyraymond, sounds reasonable. I'll see if I can make some time to migrate it to modules sometime soon. For now, I'll merge it as is, but could you please change the pull request to go against |
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hi @petergtz , i've just changed the base branch for the PR 😉 |
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I'll merge it to mainline and make a new release soon (once I have access to a computer again). |
Fix #103
The loop was writing to a variable out of the scope and the second time it goes through
if structName == "" {the condition was never false because the out of scope variable was updated with the firstIt resulted in all generated interfaces being named the same (even if the implementation was distinct for each)