🚀 go-pugleaf

In article <1995Nov23.012520.24896@il.us.swissbank.com>,
pld@syd.au.swissbank.com (Phillip Drury) writes: 
<snip>
|> Now can anyone explain to me why the compiler does not complain about
|> the second case because "const int* param" should be exactly the same
|> as "const pINT param" as there exists the typedef int *pINT.

repeat a million times: `typedef is not a macro'!!!

typedef int *pINT defines pINT to be a _pointer_ to int. It is a kind of
pointer _not_ a kind of int.

const pINT is therefore a pointer that is a constant.

const int*  is a (non-constant) pointer to a constant int.

So, const pINT is _not_ const int*.

|> 
|> Reading from the ANSI draft standard 31 October 1988 (OK its the latest I
|> can lay my hands on) it states in section A8.9
|> 
|>     "A typedef declaration attributes a type to each name among its
|> 	 declarators in the usual way. Thereafter, each such typedef name
|> 	 is syntactically equivalent to a type specifier keyword for the
|> 	 associated type"
|> 

Read again. a typedef name is syntactically equivalent to a _type specifier
keyword_ that specifies the type. It is not syntactically equivalent to the
string of tokens used to declare it.

|> This statement leads me to believe that the above behaviour exhibited by
|> these two compilers is wrong.

No: you misunderstood.

|> 
|> So what is the story. Is this a compiler bug or am I missing something
|> and if so what is the fix. I do not wish to resort to #define pINT *int,
|> if I can at all avoid it.

#define pINT *int 

is a bit hard to use anyway :-)

Given typedef int *pINT, there is no way to derive const int * from pINT.
const int * is a _very_ different object from int * there is no reason why
you should be able to get one from the other.

Cheers
Tanmoy
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