假定我有A类的B类。
A defined foo() and also, B defines foo()
我在另一个班级有办法获得B*(B点名)。 在这项职能中,我能否把A称作:foo(......),而只是利用B的点子来履行所履行的职能?
我认为,你想要做以下工作:
void fun(B* b){
b->A::foo();
}
或者,当然,你可以做的是投放类型:
((A*)b)->foo();
或以安全检查为单位的++风格:
dynamic_cast<A*>(b)->foo();
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