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C.146 Compare pb2->id() instead of pb2

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Fixes #504
This commit is contained in:
Jonathan Wakely 2016-10-10 17:31:26 +01:00
parent 173b036e51
commit 98f1e46602
1 changed files with 4 additions and 4 deletions
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CppCoreGuidelines.md
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@ -6765,14 +6765,14 @@ Consider:
cout << pb2->id(); // "D"
if (pb1->id() == pb2->id()) // *pb1 is the same type as *pb2
if (pb2 == "D") { // looks innocent
D* pd = static_cast<D*>(pb1);
// ...
if (pb2->id() == "D") { // looks innocent
D* pd = static_cast<D*>(pb1);
// ...
}
// ...
}
The result of `pb2 == "D"` is actually implementation defined.
The result of `pb2->id() == "D"` is actually implementation defined.
We added it to warn of the dangers of home-brew RTTI.
This code may work as expected for years, just to fail on a new machine, new compiler, or a new linker that does not unify character literals.