Why do we don't need to type-cast when we use new operator

[rsodimbakam]

Hi,
I am writing a sample code like this

#include <cstdlib> //declarations of malloc and free
#include <new>
#include <iostream>
using namespace std;

class C {
public:
C();
void* operator new (size_t size); //implicitly declared as a static member function
void operator delete (void *p); //implicitly declared as a static member function
};

void* C:perator new (size_t size) throw (const char *){
void * p = malloc(size);
if (p == 0) throw "allocation failure"; //instead of std::bad_alloc
return p;
}

void C:perator delete (void *p){
C* pc = static_cast<C*>(p);
free(p);
}

int main() {
C *p = new C; // calls C::new
delete p; // calls C:elete
}

Here while using the new operator or inside the new operator function, i have not mentioned any type casting to type 'C', more over i am returning a void pointer inside the operator function, in this case how the compiler is ablt to match it properly ?

Is it the C++ compiler(language) technique /Is it the new operator technique ?

[Lindley]

As another recent thread pointed out, operator new is not actually the same as the new keyword. Operator new, which is overloadable, only controls the memory-allocation portion; the constructor is then called on the allocated memory by the new keyword, and that part you can't overload.

So there's a step between the return from operator new and the return from the new keyword which is invisible to you, in which that void* is converted to a C* by the call of a constructor.

Likewise, the destructor is called before operator delete, so it is incorrect to static_cast p to a C* in that case; the C* no longer exists, just blank memory.

[rsodimbakam]

Can you also explain me about how the compiler able to allocate the required memory even though we are not using the sizeof function in case of new operator

ex:

SampleClass *sm = new SampleClass() // here i didn't used the szieof

[Paul McKenzie]

Can you also explain me about how the compiler able to allocate the required memory even though we are not using the sizeof function in case of new operator

ex:

SampleClass *sm = new SampleClass() // here i didn't used the szieof

You seem to ask a lot of questions as to why C++ is the way it is.

The creator of the C++ language wanted C++ to be a "safer" 'C', where type need not be known in many instances.

But here is the bottom line -- Every computer language has rules and syntax -- continually questioning why a language is allowed (or not allowed) to do something is really pointless.

At some point, you need to realize that C++ syntax is the way it is, and just learn how to program using the language.

Regards,

Paul McKenzie

[Richard.J]

maybe the OP's name is just a synonym for george2?