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Let's investigate closer why an tt(AirAuto) introduces i(ambiguity), when
derived from tt(Auto) and tt(Air).
itemization(
it() An tt(AirAuto) is an tt(Auto), hence a tt(Land), and hence a
tt(Vehicle).
it() However, an tt(AirAuto) is also an tt(Air), and hence a tt(Vehicle).
)
The duplication of tt(Vehicle) data is further illustrated in
fig(ambiguity).
figure(polymorphism/ambiguity)
(Duplication of a base class in multiple derivation.)
(ambiguity)
The internal organization of an tt(AirAuto) is shown in
fig(InternalOrganization)
figure(polymorphism/internal)
(Internal organization of an tt(AirAuto) object.)
(InternalOrganization)
The bf(C++) compiler will detect the ambiguity in an tt(AirAuto) object,
and will therefore fail to compile a statement like:
verb(
AirAuto cool;
cout << cool.weight() << endl;
)
The question of which member function tt(weight()) should be called,
cannot be answered by the compiler. The programmer has two possibilities to
resolve the ambiguity explicitly:
itemization(
it() First, the function call where the ambiguity occurs can be
modified. The ambiguity is resolved using the i(scope resolution operator):
verb(
// let's hope that the weight is kept in the Auto
// part of the object..
cout << cool.Auto::weight() << endl;
)
Note the position of the scope operator and the class name: before the name
of the member function itself.
it() Second, a dedicated function tt(weight()) could be created for
the class tt(AirAuto):
verb(
int AirAuto::weight() const
{
return Auto::weight();
}
)
)
The second possibility from the two above is preferable, since it relieves
the programmer who uses the class tt(AirAuto) of special precautions.
However, apart from these explicit solutions, there is a more elegant one,
discussed in the next section.
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