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copied!<p>It depends on the scale of things. For the most part it's just a couple checks and a few pointer dereferences. In most implementations, at the top of every object that has virtual functions, there is a pointer to a vtable that holds a list of pointers to all the implementations of the virtual function on that class. I would guess that most implementations would use this to either store another pointer to the type_info structure for the class.</p> <p>For example in pseudo-c++:</p> <pre><code>struct Base { virtual ~Base() {} }; struct Derived { virtual ~Derived() {} }; int main() { Base *d = new Derived(); const char *name = typeid(*d).name(); // C++ way // faked up way (this won't actually work, but gives an idea of what might be happening in some implementations). const vtable *vt = reinterpret_cast<vtable *>(d); type_info *ti = vt->typeinfo; const char *name = ProcessRawName(ti->name); } </code></pre> <p>In general the real argument against RTTI is the unmaintainability of having to modify code everywhere every time you add a new derived class. Instead of switch statements everywhere, factor those into virtual functions. This moves all the code that is different between classes into the classes themselves, so that a new derivation just needs to override all the virtual functions to become a fully functioning class. If you've ever had to hunt through a large code base for every time someone checks the type of a class and does something different, you'll quickly learn to stay away from that style of programming.</p> <p>If your compiler lets you totally turn off RTTI, the final resulting code size savings can be significant though, with such a small RAM space. The compiler needs to generate a type_info structure for every single class with a virtual function. If you turn off RTTI, all these structures do not need to be included in the executable image.</p>

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