C++ FAQ Celebrating Twenty-One Years of the C++ FAQ!!!
(Click here for a personal note from Marshall Cline.)
Section 26:
26.1 Can sizeof(char) be 2 on some machines? For example, what about double-byte characters?
26.2 What are the units of sizeof?
26.3 Whoa, but what about machines or compilers that support multibyte characters. Are you saying that a "character" and a char might be different?!?
26.4 But, but, but what about machines where a char has more than 8 bits? Surely you're not saying a C++ byte might have more than 8 bits, are you?!?
26.5 Okay, I could imagine a machine with 9-bit bytes. But surely not 16-bit bytes or 32-bit bytes, right?
26.6 I'm sooooo confused. Would you please go over the rules about bytes, chars, and characters one more time?
26.7 What is a "POD type"?
26.8 When initializing non-static data members of built-in / intrinsic / primitive types, should I use the "initialization list" or assignment?
26.9 When initializing static data members of built-in / intrinsic / primitive types, should I worry about the "static initialization order fiasco"?
26.10 Can I define an operator overload that works with built-in / intrinsic / primitive types?
26.11 When I delete an array of some built-in / intrinsic / primitive type, why can't I just say delete a instead of delete[] a?
26.12 How can I tell if an integer is a power of two without looping?
26.13 What should be returned from a function?
[26.10] Can I define an operator overload that works with built-in / intrinsic / primitive types?

No, the C++ language requires that your operator overloads take at least one operand of a "class type" or enumeration type. The C++ language will not let you define an operator all of whose operands / parameters are of primitive types.

For example, you can't define an operator== that takes two char*s and uses string comparison. That's good news because if s1 and s2 are of type char*, the expression s1 == s2 already has a well defined meaning: it compares the two pointers, not the two strings pointed to by those pointers. You shouldn't use pointers anyway. Use std::string instead of char*.

If C++ let you redefine the meaning of operators on built-in types, you wouldn't ever know what 1 + 1 is: it would depend on which headers got included and whether one of those headers redefined addition to mean, for example, subtraction.