linux-kernel - Re: Why is the kfree() argument const?

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Message-ID: <m3ve5qk8b5.fsf@maximus.localdomain>
Date:	Sat, 19 Jan 2008 00:44:46 +0100
From:	Krzysztof Halasa <khc@...waw.pl>
To:	J.A. Magallón <jamagallon@....com>
Cc:	"Linux-Kernel\, " <linux-kernel@...r.kernel.org>
Subject: Re: Why is the kfree() argument const?

"J.A. Magallón" <jamagallon@....com> writes:

> That's what __attribute__ ((pure)) is for, but if none of the
> functions is pure, the compiler can not be sure about side effects
> and can not reorder things. Don't forget that functions can do
> anything apart from mangling with their arguments.

Though it seems it could legally transform:

void kfree(const int *x);

{
int v, *ptr = malloc(sizeof(int));
*ptr = 51;
v = *ptr;
kfree(ptr);
printf("%d", v);

into:

{
int v, *ptr = malloc(sizeof(int));
*ptr = 51;
kfree(ptr);
v = *ptr;
printf("%d", v);
}

if it knows that malloc generates unaliased pointers, which seems
reasonable in general.
-- 
Krzysztof Halasa
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