JonasToth added a comment.
I see, thank you for the clarification :)
Am 15.08.2018 um 19:25 schrieb Shuai Wang via Phabricator:
> shuaiwang added inline comments.
>
> ================
> Comment at: unittests/clang-tidy/ExprMutationAnalyzerTest.cpp:410
> + match(withEnclosingCompound(declRefTo("y")), AST->getASTContext());
> + EXPECT_THAT(mutatedBy(ResultsY, AST.get()), ElementsAre("y"));
> +}
>
> ----------------
>
> JonasToth wrote:
>
>> Out of curiosity: Why is the result with `y` different from the result for
>> `x`? Both time `x` is mutated and `g()` mutates them.
>
> This is ultimately caused by not handling pointers yet.
> As soon as the address of an object is taken we assume the object is mutated.
> e.g.
>
> void f(const int*);
> void g() {
> int x;
> f(&x); // <-- address of x taken, assume mutation
> int y[10];
> f(y); // <-- address of y taken, assume mutation
> }
>
>
> And in all such cases the "mutated by" expression is the expression that
> takes the address.
>
> So back in this case, `g(x)` mutates `x` because we're assuming `g` mutates
> its argument through non-const reference. Note that the declared `g` might
> not be the one actually being called because of overload resolution, there
> could be another `void g(char(&)[8])`
> While for `g(y)` we know it's calling the `void g(char*)` so there's an
> array to pointer decay, and the decay is the point we assumed mutation not
> the function call.
>
> Repository:
>
> rCTE Clang Tools Extra
>
> https://reviews.llvm.org/D50619
Repository:
rCTE Clang Tools Extra
https://reviews.llvm.org/D50619
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