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Message-ID: <20110522160914.GR277@brightrain.aerifal.cx>
Date: Sun, 22 May 2011 12:09:14 -0400
From: Rich Felker <dalias@...ifal.cx>
To: musl@...ts.openwall.com
Subject: Re: linux/sbrk is wrong
On Sun, May 22, 2011 at 04:55:55PM +0200, Szabolcs Nagy wrote:
> linux brk syscall returns the end pointer
> so if it fails then the old one
> if it succeeds then the new one
>
> the current implementation of sbrk returns
> the end of the allocated area instead of
> the begining
IMO there's no way for correct code to use these functions anyway, but
I'll fix them anyway.
> int brk(void *end)
> {
> - return -(syscall(SYS_brk, end) == -1);
> + return -((void *)syscall(SYS_brk, end) != end);
> }
This looks fine.
> void *sbrk(ptrdiff_t inc)
> {
> - return (void *)syscall(SYS_brk, syscall(SYS_brk, 0)+inc);
> + void *p = (void *)syscall(SYS_brk, 0);
> +
> + if (inc == 0)
> + return p;
> + if ((void *)syscall(SYS_brk, p+inc) == p+inc)
This line is invalid C. p+inc is not defined since p has type void *.
The most correct would probably be to use uintptr_t internally
everywhere and just cast to void * in the return statement.
Rich
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