My C++ teacher thinks that the * operator in standard C++ is “already overloaded,” because it can mean indirection or multiplication depending on the context. He got this from C++ Primer Plus, which states:
Actually, many C++ (and C) operators already are overloaded. For example, the * operator, when applied to an address, yields the value stored at that address. But applying * to two numbers yields the product of the values. C++ uses the number and type of operands to decide which action to take. (pg 502, 5th ed)
At least one other textbook says much the same. So far as I can tell, this is not true; unary * is a different operator from binary *, and the mechanism by which the compiler disambiguates them has nothing to do with operator overloading.
Who is right?
Both are right as the question depends on context and the meaning of the word overloading.
“Overloading” can take a common meaning of “same symbol, different meaning” and allow all uses of “*” including indirection and multiplication, and any user-defined behavior.
“Overloading” can be used to apply to C++’s official operator overloading functionality, in which case indirection and multiplication are indeed different.
ADDENDUM: See Steve’s comment below, on “operator overoading” versues “token overloading”.