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From: volpe@camelback.crd.ge.com (Christopher R Volpe)
Newsgroups: comp.lang.c
Subject: Re: Assignment Ops and Side Effects
Message-ID: <18324@crdgw1.crd.ge.com>
Date: 8 Apr 91 13:14:50 GMT
References: <1991Apr3.173046.2367@mccc.edu> <1991Apr4.202314.961@csrd.uiuc.edu> <11805@dog.ee.lbl.gov> <1991Apr5.012413.24124@athena.mit.edu> <1991Apr5.234636.22610@csrd.uiuc.edu>
Sender: news@crdgw1.crd.ge.com
Reply-To: volpe@camelback.crd.ge.com (Christopher R Volpe)
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In article <1991Apr5.234636.22610@csrd.uiuc.edu>,
bliss@sp64.csrd.uiuc.edu (Brian Bliss) writes:
|>> For the compiler that thinks that sizeof(c += 1) is sizeof(int),
|>> I wonder what it thinks about sizeof(c = 1) or sizeof(c = 1.0) ?
|>(c is a char)
|>
|>The compiler is Sun 4.0's cc, and indeed, it returns 1 for both cases.
|>So indeed it is the assignment that is responsible for coercing
|>the result back to char type.
|>
|>bb

Could someone explain to me why sizeof(c=1) is 1 and not (typically) 4??
The result of the assignment operator is not an l-value, and thus undergoes
the integral promotions. The result of "c=1" in an expression context
should be promoted to int. For example, for an unprototyped function f,
f(c=1) assigns 1 to the character c, and then passes the INTEGER 1 to f.
Similarly, sizeof(c=1) should see that its operand is a non-lvalue value of
type char (the value is not evaluated) and should thus be promoted to int.
(For sizeof(c), c is an lvalue and does not undergo promotion since it is
an operand of sizeof. (Section 3.2.2.1 paragraph 2))
         
==================
Chris Volpe
G.E. Corporate R&D
volpecr@crd.ge.com