Hi Eugene,
ISO/IEC 9899 doesn't discuss this directly, but says in section 6.7.5.1:
"... const int *ptr_to_constant;
int *const constant_ptr;
The contents of any object pointed to by ptr_to_constant shall not be modified
through that pointer..."
in BN_sub, "b" is a const BIGNUM *, hence the content referenced by it may not
be modified _through b_.
The content (*b) cannot be placed in read-only storage as it is referenced, not
created, by this declaration.
This implies that it's OK to modify it _through r_.
If you were to create a const BIGNUM Z, then attempt to BN_sub(&Z, n, &Z) then
you would be violating constness by passing Z as the first (non-const)
argument. As it stands, however, the code looks fine to me.
-Yair
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED]
Sent: Wednesday, January 30, 2008 5:16 PM
To: openssl-dev@openssl.org
Subject: Re: memory corruption after usin BN_mod_inverse
Hi, Yair Elharrar!
> Sorry, I don't think that breaks any const rules.
> See explanation and example in ISO/IEC 14882 section 7.1.5.1.
First of all, OpenSSL was written in C, so ISO/IEC 14882 is not a
subject to reffer to (it is the C++ standard).
Let's see in ISO/IEC 9899 section 6.7.3:
"The implementation may place a const object that is not volatile in a
read-only region of storage." That's enough.
Then, if you look in BN_sub you'll easy understand that behavior will
be undefined if r and b point to the same object.
--
Eugene.
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