overload-resolution

I am unable to understand the output of the following C++ snippet. Should not objc.fn() call A 's fn() since B's fn() is private and should not be visible in C . However, the answer is: the call to fn() is ambiguous. How? #include<iostream> using namespace std; class A{ public: void fn() { cout << "1"; } }; class B{ void fn() { cout << "2"; } }; class C: public A, public B {}; int main(){ C objc; objc.fn(); return 0; } songyuanyao According to the book C++ Templates: The Complete Guide Appendix B.1, At a very high level, a call to a named function can be processed in the following way: The

I'm having a strange behavior with an operator overloading in C++. I have a class, and I need to check if its contents are greater or equal to a long double. I overloaded the >= operator to make this check, my declaration is as follows: bool MyClass::operator>=(long double value) const; I have to say that I also have a cast-to-long-double operator for my class, that works without exceptions only under certain conditions. Now, when I use this operator, the compiler complains that there's an ambiguous use of operator>= and the alternatives are: Mine. The built-in operator>=(long double, int) .

#include <iostream> using namespace std; template<typename T> void func(T t) { std::cout << "matched template\n"; } void func(long x) { std::cout << "matched long\n"; } int main() { func(0); } output: matched template In other cases, the non-template function is preferred when overload resolution might be ambiguous, why is this one different? §13.3.3 [over.match.best]/p1-2: 1 Define ICSi(F) as follows: (1.1) [inapplicable bullet omitted] (1.2) let ICSi(F) denote the implicit conversion sequence that converts the i-th argument in the list to the type of the i-th parameter of viable function F .