Associativity -- what is it?

[Peter J. Holsberg]

In article <230 at sdrc.UUCP> scjones at sdrc.UUCP (Larry Jones) writes:
|In article <234 at mccc.UUCP>, pjh at mccc.UUCP (Peter J. Holsberg) writes:
|> In article <224 at sdrc.UUCP> scjones at sdrc.UUCP (Larry Jones) writes:
|> |In article <226 at mccc.UUCP>, pjh at mccc.UUCP (Peter J. Holsberg) writes:
|> |> 
|> |> 	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 ????
|> |
|> |But ++ IS applied first!  The key point here is that the RESULT of postfix
|>                                                           ^^^^^^^^^^^^^^^^^
|> |++ is the value BEFORE incrementation, not that postfix ++ is somehow deferred
|>  ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|> |until later.
|> 
|> Well, that still leaves me confused.  If i has the value 7, it is 7 that
|> is added to 3, so it seems to be that the ++ *is* deferred until later. 
|> Also, ++ has higher precedence than +, so why is the incrementation
|> delayed until after the current value of i is used?
|
|The result of i++ is the value of i.  In addition, i gets incremented.  You
|can think of i++ as being like inc(&i) where inc() is defined as:
|
|   int inc(ip)
|      int *ip;
|      {
|      int j = *i;
|      *i = *i + 1;
|      return j;
|      }
|
|So, it's not the ++ operator that's defered, it's the side effect of the
     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|incrementation.  You should also be aware that the incrementation can be
|defered for a long time - it may not happen until after the assignment.
|That's why i = i++ + 2; has no defined value -- you don't know whether the
|incrementation is done before or after the assignment.
|
???? Then what does the ++ operator-sans-side effect actually do???  It
looks like it does nothing.

-- 
Peter Holsberg                  UUCP: {rutgers!}princeton!mccc!pjh
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