 Force compiler to not optimize side-effect-less statements
By : Shaggy ElBlid
Date : March 29 2020, 07:55 AM
will help you Compilers are unfortunately allowed to optimise as much as they like, even without any explicit switches, if the code behaves as if no optimisation takes place. However, you can often trick them into not doing so if you indicate that value might be used later, so I would change your code to: code :
int float_to_int(float f)
{
return static_cast<int>(f); // has no side-effects
}
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 What's the best way to make an Assembly compiler in Python available online?
By : user3551325
Date : March 29 2020, 07:55 AM
will help you The simplest solution is to put the two output files into a single zip file, send a doctype header indicating it's in zip format and send the output as raw data. The alternative is to use httprequest for the client to request each file in turn.
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Effect of Online Compiler on computer's memory
By : user590
Date : March 29 2020, 07:55 AM
hop of those help? Since everything runs on the remote machine, your local PC will not be affected by anything happening over there.
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 Why won’t C compiler give a warning when writing a function which returns nothing and has no side effect?
By : SciDat
Date : March 29 2020, 07:55 AM
this one helps. I'm not a compiler writer, so not speaking from personal experience, but I think that it's pretty rare for programmers to accidentally write this kind of function. Compiler warnings are for things that are likely to be wrong. In C, it's fairly common to use conditional compilation. Depending on compile-time option, some code is excluded. It isn't that rare to have a function like this: code :
void cleanup(void) {
#ifdef INCLUDE_FROBNICATOR
frob_close();
#endif
#ifdef INCLUDE_WIDGET_HANDLER
for (size_t i = 0; i < WIDGET_COUNT; i++)
destroy_widget(widgets[i]);
}
#endif
}
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Have issue which seems to only effect me on my computer but when I use an online compiler it works
By : user3655431
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You probably do not use Python-3.x. You probably use Python 2.7.
In 2.7, function input() attempts to evaluate your input as if it were a Python expression. It treats stop as a name of a variable that you, indeed, have never defined.
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code :
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