Parenthesis

[Kristian Kilpi]

On Sat, 23 Dec 2006 12:44:16 +0200, Wolven <rma at wolven.net> wrote:
[snip]
>>   if a == b return c;  //  errm - well...
> This one seems perfectly clear to me.  Maybe I'm reading it wrong.  To  
> me, it
> says; if a equals b then return and pass back c.  Is that not correct?

It's correct.

>>   if a == b *c++;       //  aaarrrggghhh!!!
> Likewise, this one says; if a is equal to b, add one to... pointer c?   
> Although
> I'm not positive of the meaning of the *, everything else seems crystal  
> clear.
> Again, am I not reading it correctly?  Other than my uncertainty over  
> the *, I
> don't see any ambiguity in those statements...  or any reason why C(++)
> programmers don't like them.  But since I'm not a C(++) programmer,  
> perhaps
> there's something unclear about those statements that I'm just not  
> recognizing.

The problem is how * is treaded: is it unary or binary?

I read the statement as:

if(a == b)
     *c++;

, but using arithetic precedence it will be:

if(a == b * c++);

Of course, in D you have to use {} to create an empty statement, so the  
compiler should thread it as "if(a == b) *c++;" too. (But that would add  
complexity to the compiler.)

Anyway, I think that *optional* parenthesis can make the code a bit harder  
to read. For example:

if(a && b) || c
     a = 1;

At first glance it could seem that "(a && b)" is the condition of the  
if-statement. Of course one can write unreadable code with any given  
syntax.