switch vs. initializing declarations

[John Mundt]

In article <1990Sep14.204028.21189 at ingres.Ingres.COM> jeff at ingres.com (Jeff Anton) writes:
>A few days ago, it occured to me that I didn't have a good feeling
>as to what the following code fragment which seems to be legal C
>means.  This is a retorical question and is not real world code, but I
>would like to hear from someone who has a good knowledge of the
>formal C specifications.  Please reply to me personally as I don't
>often read comp.lang.c but post to comp.lang.c if you wish.
>
>main(argc, argv)
>int	argc;
>char	*argv[];
>{
>	switch (argc) {
>		int	v = 1;
>
>	default:
>		v += 5;
>	case 1:
>		printf("%d\n", v);
>	}
>	return 0;
>}
>
>
>The ambiguity is whether or not 'v' should be initialized or not.
>All compilers I've tested recognize the declaration but do not
>do the initialization.  Some report line 6 statement not reached when
>clearly the statement does have the declaritoy effect....

I'm surprised it compiles, but it does.  Line 6 is not reached because
it is not within any of the case statements.  Therefore, there is no
argument you can give to argc which will reach the "case" of 
int	v = 1; so it is never executed.

It runs probably because most compilers assign a type of int to
variables and functions not specifically declared.  lint
has this to say about the program:

warning: statement not reached
    (6)  	
-- 
---------------------
john at admctr.chi.il.us
John Mundt   Teachers' Aide, Inc.  P.O. Box 1666,  Highland Park, IL
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