@tom_forsyth @TheJare @vgebler It's basically a Compiler Who Cried Wolf problem :P
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Replying to @cmuratori
@cmuratori@TheJare@vgebler For every "pointless" warning, I can usually think of a bug it would have caught, making it an overall win.1 reply 0 retweets 0 likes -
Replying to @tom_forsyth
@tom_forsyth@TheJare@vgebler const warnings for strings, while(1), etc. There are no bugs that these can catch.1 reply 0 retweets 0 likes -
Replying to @cmuratori
@cmuratori@TheJare@vgebler "Expression is constant" is useful, it just needs nerfing when it's deliberate.1 reply 0 retweets 0 likes -
Replying to @tom_forsyth
@tom_forsyth@TheJare@vgebler But while(1) is always deliberate, this is my point. Thus that whole series of warnings was illegitimate.1 reply 0 retweets 0 likes -
Replying to @cmuratori
@tom_forsyth@TheJare@vgebler while(var that doesn't happen to change) is very different from while(1).2 replies 0 retweets 0 likes -
Replying to @cmuratori
@tom_forsyth@TheJare@vgebler It is not like an unused variable warning, where it _might_ not have been your intention.1 reply 0 retweets 0 likes -
Replying to @cmuratori
@cmuratori@TheJare@vgebler It's intended to catch if(x<0) where x is unsigned for example. So the warning has a use, just not that case.1 reply 0 retweets 0 likes -
Replying to @tom_forsyth
@tom_forsyth@TheJare@vgebler You are listening to me. I'm saying that the compiler warns on things that are not warnings.1 reply 0 retweets 0 likes -
Replying to @cmuratori
@tom_forsyth@TheJare@vgebler That is true. Just because a bad compiler implementation treats those things as the same doesn't change it.1 reply 0 retweets 0 likes
@tom_forsyth @TheJare @vgebler You are explaining _why_ the compiler produces the erroneous warning. But that is irrelevant.
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