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author | David Majnemer <david.majnemer@gmail.com> | 2014-08-08 00:10:39 +0000 |
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committer | David Majnemer <david.majnemer@gmail.com> | 2014-08-08 00:10:39 +0000 |
commit | d96b99740d1f2c47941e589d9719016388ee6453 (patch) | |
tree | 1948bfebd62c54547ae481c4ad1b81b622126bcc /llvm/lib/TableGen/TGParser.cpp | |
parent | 7d498629f1e038d2642116b40a04f4acf6532a26 (diff) | |
download | llvm-d96b99740d1f2c47941e589d9719016388ee6453.zip llvm-d96b99740d1f2c47941e589d9719016388ee6453.tar.gz llvm-d96b99740d1f2c47941e589d9719016388ee6453.tar.bz2 |
MS ABI: Don't force bases to have an inheritance model
Previously, assigning an inheritance model to a derived class would
trigger further assiginments to the various bases of the class. This
was done to fix a bug where we couldn't handle an implicit
base-to-derived conversion for pointers-to-members when the conversion
was ambiguous at an earlier point.
However, this is not how the MS scheme works. Instead, assign
inheritance models to *just* the class which owns to declaration we
ended up referencing.
N.B. This result is surprising in many ways. It means that it is
possible for a base to have a "larger" inheritance model than it's
derived classes. It also means that bases in the conversion path do not
get assigned a model.
struct A { void f(); void f(int); };
struct B : A {};
struct C : B {};
void f() { void (C::*x)() = &A::f; }
We can only begin to assign an inheritance model *after* we've seen the
address-of but *before* we've done the implicit conversion the more
derived pointer-to-member type. After that point, both 'A' and 'C' will
have an inheritance model but 'B' will not. Surprising, right?
llvm-svn: 215174
Diffstat (limited to 'llvm/lib/TableGen/TGParser.cpp')
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