Example Program: a[i]=b[i++] non-portability

[Joseph S. D. Yao]

In article <359 at g.cs.cmu.edu> ckk at g.cs.cmu.edu (Chris Koenigsberg) writes:
>#include <stdio.h>
>main(argc, argv)
>	{char a[10];static char b[]="abcdefghij";int i,j;
>	printf("a =%s, b=%s\n", a, b);
>	printf("Looping over i...");
>	while(b[i] != 0) a[i] = b[i++];
>	printf("New a =");
>	for(i=0;i<=10;printf("%c",a[i++]));printf(", b=%s\n",b);
>	fflush(stdout);
>	}
>On a sun, the output looks like:
>a =, b=abcdefghij
>Looping over i...New a =abcdefghij^@, b=abcdefghij
>And on the IBM RT PC, the output looks like:
>a =, b=abcdefghij
>Looping over i...New a =^@abcdefghij, b=abcdefghij
>Note: The "^@" was actually a null character, in the output.  
>On the sun, a[0] gets b[0], but on the RT PC, a[1] gets b[0] and a[0] is
>a null.

Chris, run lint on this.  One problem is that I IS NOT INITIALISED!
Then, you index into a[10] both in the assignment and in the printing
loops.  Elements a[0-9] exist: a[10] is an addressing error.  On the
8086, what model was this compiled under?  You were lucky that this
ran.  (Yes, if core is zeroed first i will always be 0.  What happens
when you try to make this code a re-entrant subroutine?)

As for the main point, this is working OK: the code does what people
were saying it should do.  I don't have your original message here:
is this  r e a l l y  exactly what got folk upset?