Re: [Qemu-devel] [PATCH] use qemu_malloc and friends consistently

[jcd@tribudubois.net]

----- "Markus Armbruster" <armbru@redhat.com> a écrit :
> 
> No.  malloc() implements the same, well-known spec everywhere.  Some
> people don't understand or don't like the spec, but that doesn't make
> implementations of the spec incompatible.  Here's what you can expect
> from malloc():
> 
> * After p = malloc(SIZE), you can access p[i] for 0 <= i < SIZE.  It
>   follows that you must not dereference p at all if SIZE == 0.
> 
> * If malloc(SIZE) returns a null pointer, and SIZE != 0, then
> malloc()
>   failed.
> 
> * Any result of malloc() can safely be passed to free().
> 
> A wrapper like qemu_malloc() can implement a different spec, of
> course.
> But only if the spec is compatible, the wrapper can be used as a
> drop-in
> replacement.  Making it gratuitously incompatible is foolish, because
> it
> *breaks* code that works just fine with malloc().
> 
> We already agreed on one compatible difference: qemu_malloc() shall
> not
> fail, but just terminate the program.
> 
> We're debating another difference, namely what qemu_malloc(0) shall
> do.
> Two fine compatible options have been proposed here:
> 
> 1. qemu_malloc(0) shall return a null pointer.
> 
> 2. qemu_malloc(0) shall return a non-null pointer that can be passed
> to
>    qemu_free()
> 
> The one that got committed is incompatible.

Thanks for the clarification Markus.

JC