Re: [tlug] C puzzle

[bruno raoult <br@example.com>]

Exactly, it looks correct to me: If I want to get the smallest C
program to crash, it would certainly not be "correct" in some
manner, but this is what I wish!
It is probably more correct than a "correct" program which crashes
when it should not :-)

It means we got 2 solutions for this 1-char output problem :-)

br.

PS. BTW, what is this "smallest C program", which compiles with
no options?


On Tue, 2002-07-30 at 11:10, Josh Glover wrote:
> > I don't think this one satisfies the original request to turn the
> > sample into a "correct C program". The original code is much more
> > correct than your solution:)
> 
> No, my friend. Whether C is "correct" or not is dependent on only one 
> thing: compiling and doing what it is supposed to. Macho C hackers 
> cackle with glee over this clever use of pointer arithmatic. :)
> 
> > Nevertheless it works (despite of gcc complaints). I think it's
> > because the pointer to int is unsigned, so negating it results in a
> > big positive number.
> 
> You would seem to be right:
> 
> (gdb) l
> 1
> 2       main() {
> 3         int*i, n=20;
> 4         for(i = 0 ; i < n ; i--)
> 5           printf("-");
> 6       }
> (gdb) b 3
> Breakpoint 1 at 0x8048466: file test.c, line 3.
> (gdb) r
> Starting program: /home/jmglov/test
> 
> Breakpoint 1, main () at test.c:3
> 3         int*i, n=20;
> (gdb) n
> 4         for(i = 0 ; i < n ; i--)
> (gdb) p i
> $1 = (int *) 0x804961c
> (gdb) n
> 5           printf("-");
> (gdb) p i
> $2 = (int *) 0x0
> (gdb) p n
> $3 = 20
> (gdb) p i < n
> $4 = 1
> (gdb) n
> 4         for(i = 0 ; i < n ; i--)
> (gdb) p i
> $5 = (int *) 0x0
> (gdb) n
> 6       }
> (gdb) p i
> $6 = (int *) 0xfffffffc
> (gdb) p i < n
> $7 = 0


-- 
Any sufficiently advanced bug is indistinguishable from a feature.
                -- Rich Kulawiec

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