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From: crowl@cs.rochester.edu (Lawrence Crowl)
Newsgroups: comp.lang.c
Subject: Re: Possible C Anomaly
Message-ID: <2174@sol.ARPA>
Date: Sat, 12-Sep-87 11:22:11 EDT
Article-I.D.: sol.2174
Posted: Sat Sep 12 11:22:11 1987
Date-Received: Sun, 13-Sep-87 09:43:17 EDT
References: <9236@brl-adm.ARPA>
Reply-To: crowl@cs.rochester.edu (Lawrence Crowl)
Organization: U of Rochester, CS Dept, Rochester, NY
Lines: 28

In article <9236@brl-adm.ARPA> gamma@EDN-UNIX.arpa (W. J. Showalter) writes:
]
]/* Example #1 */            k = (k > 1) ? k-- : 1;
]
]/* Example #2 */            if (k > 1) k = k--;
]                                  else k = 1;
]
]The problem is is that the two examples behave differently.  In the first
]example k does not change values.  It remains equal to 10 indefinitely.
]
]In the second example k does change like one would expect.
]
]I generated assembler listings for the two examples and discovered that
]example 1 makes used of an interim register to house the value of k
]BEFORE decrementing.  It eventually moves this value back to k, replacing
]the decremented value with the original.

The order of argument evaluation within expressions is undefined in C.  The
order of expression evaluation only matters when the expression has side
effects on other parts of the expression, as yours does.  Because of the
undefined order, the compiler may choose to execute either k = k or k-- first.
In the first example, it decrements k first and then assigns the old value of
k to k.  In the second example, it assigns the old value of k to k and then
decrements k.  In short, the code is wrong, not the compilers.
-- 
  Lawrence Crowl		716-275-9499	University of Rochester
		     crowl@cs.rochester.arpa	Computer Science Department
 ...!{allegra,decvax,seismo}!rochester!crowl	Rochester, New York,  14627