C++ – Do I need to use dynamic_cast when calling a function that accepts the base class

c++-clidynamic-castfunction

I have some classes like this:

interface class IA
{
};

interface class IB
{
};

public ref class C : public IA, public IB
{
};

public ref class D
{
    void DoSomething(IA^ aaa)
    {
    }

    void Run()
    {
        C^ bob = gcnew C();
        DoSomething(dynamic_cast<IA^>(bob));    // #1
        DoSomething(bob);           // #2
    }
};

At the moment I always try to use dynamic casting when calling such a function, (the #1 above).
However it does make the code quite ugly, so I want to know if it is actually necessary.

Do you use dynamic_cast in this way? If so what is the main reason?

Best Answer

In standard C++, you use dynamic_cast to walk down the hierarchy, not up. In this case, you'd use it to try and convert an IA or IB into a C:

IA^ temp = /* get a C in some way. */;
C^ tempC = dynamic_cast<C^>(temp);

Related Solutions

C++ – Safely checking the type of a variable

I've had to do similar things when loading C++ DLLs in apps written in languages that only support a C interface. Here is a solution that will give you an immediate error if an unexpected object type was passed in. This can make things much easier to diagnose when something goes wrong.

The trick is that every class that you pass out as a handle has to inherit from a common base class.

#include <stdexcept>
#include <typeinfo>
#include <string>
#include <iostream>
using namespace std;


// Any class that needs to be passed out as a handle must inherit from this class.
// Use virtual inheritance if needed in multiple inheritance situations.
class Base
{

public:
    virtual ~Base() {} // Ensure a v-table exists for RTTI/dynamic_cast to work
};


class ClassA : public Base
{

};

class ClassB : public Base
{

};

class ClassC
{
public:
    virtual ~ClassC() {}
};

// Convert a pointer to a long handle.  Always use this function
// to pass handles to outside code.  It ensures that T does derive
// from Base, and that things work properly in a multiple inheritance
// situation.
template <typename T>
long pointer_to_handle_cast(T ptr)
{
    return reinterpret_cast<long>(static_cast<Base*>(ptr));
}

// Convert a long handle back to a pointer.  This makes sure at
// compile time that T does derive from Base.  Throws an exception
// if handle is NULL, or a pointer to a non-rtti object, or a pointer
// to a class not convertable to T.
template <typename T>
T safe_handle_cast(long handle)
{
    if (handle == NULL)
        throw invalid_argument(string("Error casting null pointer to ") + (typeid(T).name()));

    Base *base = static_cast<T>(NULL); // Check at compile time that T converts to a Base *
    base = reinterpret_cast<Base *>(handle);
    T result = NULL;

    try
    {
        result = dynamic_cast<T>(base);
    }
    catch(__non_rtti_object &)
    {
        throw invalid_argument(string("Error casting non-rtti object to ") + (typeid(T).name()));
    }

    if (!result)
        throw invalid_argument(string("Error casting pointer to ") + typeid(*base).name() + " to " + (typeid(T).name()));

    return result;
}

int main()
{
    ClassA *a = new ClassA();
    ClassB *b = new ClassB();
    ClassC *c = new ClassC();
    long d = 0; 


    long ahandle = pointer_to_handle_cast(a);
    long bhandle = pointer_to_handle_cast(b);
    // long chandle = pointer_to_handle_cast(c); //Won't compile
    long chandle = reinterpret_cast<long>(c);
    // long dhandle = pointer_to_handle_cast(&d); Won't compile
    long dhandle = reinterpret_cast<long>(&d);

    // send handle to library
    //...
    // get handle back
    try
    {
        a = safe_handle_cast<ClassA *>(ahandle);
        //a = safe_handle_cast<ClassA *>(bhandle); // fails at runtime
        //a = safe_handle_cast<ClassA *>(chandle); // fails at runtime
        //a = safe_handle_cast<ClassA *>(dhandle); // fails at runtime
        //a = safe_handle_cast<ClassA *>(NULL); // fails at runtime
        //c = safe_handle_cast<ClassC *>(chandle); // Won't compile
    }
    catch (invalid_argument &ex)
    {
        cout << ex.what() << endl;
    }

    return 0;
}

Javascript – How to get the function name from within that function

In ES6, you can just use myFunction.name.

Note: Beware that some JS minifiers might throw away function names, to compress better; you may need to tweak their settings to avoid that.

In ES5, the best thing to do is:

function functionName(fun) {
  var ret = fun.toString();
  ret = ret.substr('function '.length);
  ret = ret.substr(0, ret.indexOf('('));
  return ret;
}

Using Function.caller is non-standard. Function.caller and arguments.callee are both forbidden in strict mode.

Edit: nus's regex based answer below achieves the same thing, but has better performance!

Best Answer

Related Solutions

C++ – Safely checking the type of a variable

Javascript – How to get the function name from within that function

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