Path: utzoo!mnetor!uunet!seismo!sundc!pitstop!sun!decwrl!labrea!aurora!amelia!ames!killer!fmsrl7!eecae!upba!cory
From: cory@upba.UUCP
Newsgroups: comp.lang.c
Subject: Re: Associativity -- what is it?
Message-ID: <51200001@upba>
Date: 25 Feb 88 17:13:00 GMT
References: <226@mccc.UUCP>
Lines: 37
Nf-ID: #R:mccc.UUCP:226:upba:51200001:000:1707
Nf-From: upba.UUCP!cory    Feb 25 11:13:00 1988


(Sorry for posting instead of mailing, I couldn't find a path to mccc)

> 	x = 3 * i ++;
> 
> Book says that ++ has a higher precedence than *, and that ++
> associates from R->L.  That makes me think that ++ should be applied
> first, but I know it isn't.  But ????

I can't really give you an understanding of associativity, as I don't know
what you are missunderstanding.  I could quote the definition, but I don't
think that would solve your problem.  I CAN try to help you understand the
concept of precedence in expression evaluations.  You seem to be confused
on the use of "var ++".  First, think of precedence as just putting a lot
of parentheses everywhere.  You know that ++ has a higher precedence than
*, so lets rewrite your example with parentheses to show the precedence:
	 	x = ( 3 * ( i ++ ) );
Now lets evaluate the expression as it is written.  First the inner-most
parentheses are evaluated:
	EXP=	( i ++ )
This reads "get the value of i, increment i, and return the pre-increment
value".  The value of this part of the expression is the before-increment
value of i.  Now lets continue to the next level of parentheses:
	EXP=	( 3 * EXP )
This says multiply 3 times the value of the inner expression. (which was
the pre-increment value of i)  Finally:
	 	x = ( 3 * ( i ++ ) );
increments i by 1 unit of i's variable type and then assigns x the value
of 3 times the pre-increment value of i. (guzzuntite)

Hope this has been some help.

					-Cory

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