lclint-interest message 198
From 100566.1506@csi.com Fri Feb 27 14:10:45 1998
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Date: Fri, 27 Feb 1998 14:17:20 +0100 (CET)
From: Hermann Kleier <100566.1506@csi.com>
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Reply-To: Hermann Kleier <100566.1506@csi.com>
To: David Evans
Cc: lclint-interest@larch.lcs.mit.edu, HermannK@compuserve.com, weco@csi.com
Subject: Re: signal function interface
In-Reply-To: <9802261752.AA24516@larch.lcs.mit.edu>
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Dave,
On Thu, 26 Feb 1998, David Evans wrote:
> > In contradiction to LcLint, Flexelint 7.5 did not report that `fa' in line 13
> > has not been initialized. I feel that `fa' is an undefined variable there and
> > NOT an external function. Is this a bug in FlexeLint?
>
> I didn't mean to suggest that it is okay for LCLint to typecheck
> standard C programs differently than C does. But, I do need to
> understand what the rules are before I can implement this correctly.
I opened another thread of discussion in my reply. I know that this is not
a good style. But I feel that this thread should be very tiny. In order
to understand your test program I passed it through my compilers and lints. And
then I stumbled over a diagnostic from LcLint that FlexeLint did not report. In
my opinion LcLint is right and I just wanted to know whether there is somebody
out there who argues against it.
> As far as I can find, the standard doesn't have any explanation of
> implicit function pointers, it just seems to be something
> implementations have adopted because of traditions. I must be missing
> something, but nothing I can find in the standard indicates that func1
> and func1 * should be the same type. In fact, it seems to directly
> contradict what I read in Sec 3.5.4.1: "For two pointer types to be
> compatible, both shall be identically qualified and both shall be
> pointers to compatible types."
I must admit that I do not have an access to the CURRENT definition of ANSI C.
There are some hints in `Harbison, Samuel P.; Steele, Guy S.: C AReference
Manual. 1987' (I have got a newer ed. of this book and will verify the info
there):
6.3.2 The Assignment Conversions (pg 135): A Right Side Type ``function'' is
converted to a Left Side Type ``pointer to function''.
This is just an intermediate step while doing any assignments. But any
assignment of function names has to come through this needle hole.
6.3.3 The Usual Unary Conversions (pg 135): ``function returning T'' is
converted to ``pointer to function returning T''.
This means that in respect to unary operators there are no ``function'' types.
There are only ``pointer to function'' types .
6.2.7 Conversion to Pointer Types (pg 133): An expression of the type
``function returning T'' for some type T is converted to a value of type
``pointer to function T'' by substituting a pointer to the function for the
function itself. The conversion occurs implicitly in all context except the
function expression in a function call.
And more explicitly:
7.3.1 Names (pg 144): The name of a function evaluates to that function
evaluates to that function; it is not an lvalue. In contexts where the
result is subject to the usual conversions, the function value is
immediately converted to a pointer to the function. This occurs in all
contexts but two: as the argument to the sizeof operator, where the
function is illegal, and as the function in a function call expression, in
which case the function itself is desired, and not a pointer to it.
To say it with my own words: Don't care about whether a variable is typed
``function returning T'' or ``pointer to function returning T''. (Almost)
anything you are going to do with it, hides the difference. The only exception
is the sizeof-operator. Lets have a look at an example:
#include
typedef void Fun (const char *);
typedef void (*FunPtr) (const char *);
int main (void);
static Fun say; // declare say: value is assigned later
int main ()
{
FunPtr sayptr; // declare sayptr
sayptr = say; // assignment o.k. This does not work
// the other way around, because say is
// no lvalue
say ("say\n"); // standard call
sayptr ("sayptr\n"); // a pointer is mapped to the function
// itself
(*say) ("(*say)\n"); // The unary operator `*' causes say to
// be converted to a pointer. This
// conversion is reversed by the `*'.
(*sayptr) ("(*sayptr)\n"); // standard call
printf ("%d\n",
(int) sizeof sayptr); // yields the size of the pointer
printf ("%d\n",
(int) sizeof say); // ILLEGAL
return 0;
}
static void say (const char *msg) // the ONLY way to assign a value
// to a variable typed ``function
// returning T''.
{
(void) fputs (msg, stdout);
}
I think the comments stand for themselves. The only illegal point is the
`sizeof' operator applied to `say' (a variable typed ``function returning T''.
FlexeLint (Ver 7.50d) reports
(int) sizeof say); // ILLEGAL
t.c 27 Error 139: Can't take sizeof function
which is o.k. GNU's cc (version 2.7.2.1) reports
t.c: In function `main':
t.c:27: warning: sizeof applied to a function type
Again, this is o.k. IMHO LCLint (2.3i --- 15 Sep 97) incorrectly reports:
t.c:31,13: Variable say redefined
A function or variable is redefined. One of the declarations should use
extern. (-redef will suppress message)
t.c:7,12: Previous definition of say
t.c:31,13: Variable say inconsistently redeclared as function
A function, variable or constant is redefined with a different type.
(-incondefs will suppress message)
t.c:7,12: Previous definition of say: Fun
t.c:31,13: File static function say declared but not used
A function is declared but not used. Use /*@unused@*/ in front of function
header to suppress message. (-fcnuse will suppress message)
t.c:35,1: Definition of say
But LcLint does not worry about the error in line 27.
Again, I feel that the definition of C is unnecessary complex here. And
searching for a formal specification of this setup is hard.
Hermann
David
Evans
University of Virginia, Computer Science
evans@cs.virginia.edu