The signature for the new operator is:
void* operator new(size_t count)
There is a white space between the word "operator" and the word "new". This is:
Different from all other operator signatures (besides new, delete and their array counterparts). for example:
T& operator=(const T& other)
Does not comply with function name rules of C++ which do not allow white spaces.
I assume that "operator" is a keyword in this case... But if this is correct, why isn't this the case in all other operator function signatures?
Is there an explanation for the inconsistency?
The C++ grammar is actually written in terms of tokens, which is consistent between all types of operator definitions. operators are special-purpose and fixed in terms of what tokens can follow after the operator keyword, but the whitespace that occurs is not mandated anywhere in the standard.
As far as the C++ grammar is concerned, there is no difference between the following signatures:
auto operator+=(const Foo&) -> Foo&
auto operator +=(const Foo&) -> Foo&
auto operator += (const Foo&) -> Foo&
auto
operator
+=
(const Foo&) -> Foo&
operator new is no more, or less, consistent here. The only deviation here is that some operators allow no space after the operator token, such as operator= -- and the reason for this is the way that tokens are broken up in the C++ grammar.
Most tokens are alphanumeric sequences that break on special characters, such as whitespace or +,=, etc. In this case, operator= breaks into two tokens -- operator and =, which form the operator.
This also follows that some operators are also compounds of multiple tokens, such as operator new[] -- which can also be written operator new [], since it is comprised of 3 tokens: operator, new, and [], as evident in the footnote from [over.oper.general]/1:
[Note 1: The operators new[], delete[], (), and [] are formed from more than one token. The latter two operators are function call and subscripting. — end note]