Pretty basic code:Which outputs:As you can see, b and bConst seem to be treated using the same value - i.e. it prints for both the same 0.057762265046662104872599030613855575211346149444580078125 value. So I guess they are "stored" both the same. The only difference is that b is not const.
Please note: b is of const T& and ++b is ok!Why is const T& not sure to be const?Welcome to const and reference collapsing. When you have const T& the reference gets applied to T and so does the const. You call g like
The "const" here is the cause of the compilation problem. However, having implemented an STL tree myself, I can't understand why.
Is this code UB?In other words, is the (temporary) object originally const? I've looked through the standard but cannot find an answer so would appreciate quotations to relevant sections.
In the following code, I throw an int, catch it as const int&, re-throw it and catch it again catch it as an int&.
I'm still confusing where to place const in pointers with more than one indirection. Can someone clarify?
I have a class Stuff, with two functions foo (const and non-const):Here's what I am trying to do:If the stuff was const, call const foo in the destructor of Stuff.
I'm working on learning C++ with Stroustrup's (Programming Principles & Practice Using C++) book. In an exercise we define a simple struct:
C++ provides implicit conversion from T * to const T *.If I use T * within a container class now, like in vector<T *>, then there is of course no implicit conversion to vector<const T *> anymore.
C++11 has given us std::add_const; with C++17, we have a new structure - std::as_const(). The former just tacks a const before the type you provide it with. The second one is a proper (template of a) function, not type trait, which seems to do the same - except for when...