[ruby-talk:16348] Re: Little bug in the manual

----- Original Message -----
From: Mathieu Bouchard <[e-mail]>
To: ruby-talk ML <[e-mail]>; teespy <[e-mail]>
Cc: ruby-talk ML <[e-mail]>
Sent: Thursday, June 07, 2001 7:30 PM
Subject: [ruby-talk:16313] Re: Little bug in the manual


> > I think there is a little error in ruby's guide
> > ( www.ruby-lang.org/en/pdf-doc/rubyguide_A4.pdf )
> > Look at page 16, the C example of a loop, quoting:
> > char *str;
> > for (str = "abcdefg"; *str != '\0'; str++) {
> > /* process a character here */
> > }
> > I think  'str = "abcdefg"' isn't legal, and should be replaced with
> > strcpy(str, 'abcdefg');
> > That's it, nothing important, but may be worth fixing in the next
release of
> > the guide.
>
> Not only str="abcdefg" is legal and working, but your replacement has
> several errors in it.
>
> Please learn the difference between a character literal and a string
> literal; and then learn about initializing all your variables; then learn
> about the use of memory allocation and deallocation; then learn how you
> can fix your code; then learn why the original code can be right too.
>
> That's it, nothing important, but may be worth learning for your next
> post about C code.

Let us give him a little assistance here.

When you declare a pointer in C, you get only a pointer. There is no memory
allocated for the thing pointed to. (It is truly frustrating how many people
don't
seem to know this -- it seems like a C newbie issue, but I have been
frightened by "oldbies" (?) who didn't grasp it...)

    char *str;
    strcpy(str,"hello");    /* Coding error! Probable crash, now or later.
*/

This is better:

    char *str;
    str = (char *) malloc(10);
    strcpy(str,"hello");    /* No problem. */

Of course, the use of malloc() implies the later use of free(), or you will
have a memory leak.

A string constant is stored internally as a pointer to a character string.
So it
is legal to assign the pointer value to a pointer variable. No malloc() is
required, since the memory already exists.

    char *str;
    str = "hello";   /* Memory for constant allocated at compile time. */

Does that help some?

Cheers,
Hal