Is “const LPVOID” equivalent to “void * const”?

早过忘川 提交于 2019-11-29 02:01:30

A typedef-name denotes a type, and not a sequence of tokens (as does a macro). In your case, LPVOID denotes the type also denoted by the token sequence void *. So the diagram looks like

// [...] is the type entity, which we cannot express directly.
LPVOID => [void *] 

Semantically if you specify the type const LPVOID, you get the following diagram (the brackets around the specifiers mean "the type denoted by the specifier"):

// equivalent (think of "const [int]" and "[int] const"):
const LPVOID <=> LPVOID const =>  const [void *] <=> [void *] const  
                              =>  ["const qualified void-pointer"]

It's not the same thing as the token sequence const void * - because this one would not denote a const qualified pointer type, but rather a pointer to a const qualified type (the thing pointed to would be const).

Syntactically a parameter declaration has the following (simplified) form:

declaration-specifiers declarator

The declaration-specifiers in case of const void *p are const void - so the base-type of *p is a const qualified void, but the pointer itself is not qualified. In case of const LPVOID p however the declaration-specifiers specify a const qualified LPVOID - which means the pointer type itself is qualified, making the parameter declaration identical to void *const p.

LPVOID is far generic pointer, which is already long time the same as normal generic pointer (it was different on old 16bit platforms).

MGC
void* const x = 0;
x = 0; // this line will not compile - u cannot change x, only what it points to
x->NonConstMethod(); // will compile
const void* y = 0;
y = 0; // this line will compile - u can change y, but not what it points to
y->NonConstMethod(); // will not compile
const void* const z = 0; // u cannot change z or what it points to
// btw, the type of the 'this' pointer is "ClassName* const this;"
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