Path: utzoo!attcan!uunet!aplcen!haven!mimsy!chris
From: chris@mimsy.umd.edu (Chris Torek)
Newsgroups: comp.lang.c
Subject: Re: typedef-ing an array
Message-ID: <25247@mimsy.umd.edu>
Date: 2 Jul 90 23:58:01 GMT
References: <78627@srcsip.UUCP> <78633@srcsip.UUCP>
Organization: U of Maryland, Dept. of Computer Science, Coll. Pk., MD 20742
Lines: 72

In article <78633@srcsip.UUCP> pclark@SRC.Honeywell.COM (Peter Clark) writes:
>> ... typedef char foo[24]; ...
>>Is there anyway to typedef an array (of a specific length)? Obviously, the
>>above declaration doesn't do it, or I wouldn't be asking. ...

The typedef above *does* do it.  After

	typedef char foo[24];

`foo' is an alias for a type that is `array 24 of char' and has size
24*sizeof(char)==24.  Thus, e.g.,

	printf("%d\n", sizeof(foo));

must print 42.

>What I meant was, why doesn't this work:

>typedef char foo[29];
>/* a string with 28 chars, add one for the NULL byte */
>foo bar = "Hello World, this is my test";
>void
>  main()
>{
>  bar = "Silly old me";
>  printf("%s\n",bar);
>}

>The initializer works fine, but the assignment inside main() doesn't.

The first error is `void main': it must (yes must) be `int main', even
if it never returns.  It may have either 0 arguments or two (int argc,
char *argv).

The second, more important error is the assignment inside main.  `bar'
is an object of type `array 29 of char'; a quoted string is an object
of type `array N of char' where N is one more than the number of characters
in the string.  `Silly old me' is 12 characters, so we have

	<object, array 29 of char, bar> =
		<object, array 13 of char, "Silly old me\0">;

The thing on the right hand side is in a value context, so it undergoes
The Famous Rule [repeat: in any value context an object of type `array
N of T' becomes a value of type `pointer to T' whose value is the address
of the first (0th) element of the array]:

	<object, array 29 of char, bar> =
		<value, pointer to char, 'S' in "Silly old me\0">

We now have a type mismatch---an array on the left and a pointer on the
right---as well as another constraint violation: arrays are not modifiable
lvalues and may not appear on the left of an assignment operator.

One obvious way to change it runs into another problem:

 1>	int main() {
 2>		foo local = "Silly old me";
		...

Now `local' is a local variable and the string on the right is an
initializer.  Line 2 here is NOT an executable statement, but rather a
declaration with an initializer.  Under ANSI C, aggregate initializers
are always permitted provided that the initializer is constant (which
this is).  (The string on the right of the equal sign is not in a value
context; it is in an initializer context.) If you hand this to a
Classic C compiler, however, you will get a complaint, because Classic
C allowed aggregate initializers only when the object being initialized
was global or static.
-- 
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7163)
Domain:	chris@cs.umd.edu	Path:	uunet!mimsy!chris