Re: difference between typedef and #define

On Thu, 30 Jun 2005 18:06:41 -0500, Drew Myers wrote:

> But if I, without using typedef or #define write:
>
> int* i,j,k; // * next to int and not next to i
>
> all of the variables are int*

No, only i is int*. The rest are int.

For background on this, see:

Is ``int* p;'' right or is ``int *p;'' right?
http://www.research.att.com/~bs/bs_faq2.html#whitespace

> whereas:
>
> int *i,j,k; // * next to i and not next to int
>
> produces the results you explain. Why does typedef/#define
> behave differently than typing it out as I did above?

Macros do pure textual substitution. So, given:

#define T int*
T i, j, k;

The preprocessor turns this into:

int* i, j, k;

The typedef mechanism is different. It doesn't do textual substitution but
instead creates synonyms for types. So, given:

typedef int* T;
T i, j, k;

This results in all the variables having the type T, which is int*. To see
another difference, consider:

const T x;
T const x;

The const (or volatile) applies to the variable x in the first case as well
as the second, not to the referent (i.e. *x), and that's another difference
between the macro T and typedef T. So, when you write a pointer or
reference typedef, you should also provide const/volatile forms as
necessary, and your users will have to learn to use LPCSTR instead of
"const LPSTR" or "LPSTR const", to use a Windows example.

--
Doug Harrison
Microsoft MVP - Visual C++
.

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