new operator for memory allocation on heap

Question

I was looking at the signature of new operator. Which is:

void* operator new (std::size_t size) throw (std::bad_alloc);

But when we use thi

Answer 1

You confuse new expression with operator new() function. When the former is compiled the compiler among other stuff generates a call to operator new() function and passes size enough to hold the type mentioned in the new expression and then a pointer of that type is returned.