may help you . How does const work? It is always applied to what comes right before it. However, when there is nothing before, it applies to what comes right after it. In other words, const Type& is the same as Type const&. Now, do not ever use preprocessor macros to define type aliases. There are two ways of doing that in C++: using and typedef. The fact you used a macro is the reason why you observe this peculiar behaviour. The code that fails to compile is expanded to the following:
void fun(const C*& p); // The pointee is const, but the pointer itself is not
void fun(C* const& p); // The pointee is not const, but the pointer itself is
Any of those help Looks like an error in the book ... you definitely must name a type or aliased type (i.e., a typedef) since C++ is a strongly-typed languages. Here is what the C++03 specification states on objects, declarations and definitions:
const keyword in return type of function declaration
I hope this helps you . I have written following 3 functions in C++. Kindly explain me how all return types are different? And how the return values will be stored in Memory? I know const keyword applies to whatever is on immediate left but i need more explanation.
const int* sample1();
int const* sample3();
int* const sample2();
Passing lambda declared using auto-keyword by non-const reference as argument to std::function parameter type
With these it helps In your fizz(g1) case, g1 is already a std::function. Therefore, there is no need for an implicit conversion. So fizz's parameter is a non-const reference to g1. In fizz(g2), g2 is not a std::function. Therefore, the compiler must perform an implicit conversion. And the implicit conversion returns a temporary.
const char* classname::getname( ) const ? whats the meaning of first const keyword?