On Fri, Jul 5, 2024 at 3:05 PM Ruediger Pluem <rpl...@apache.org> wrote:
>
>
>
> On 7/5/24 2:14 PM, Ruediger Pluem wrote:
> >
> >
> > On 7/5/24 2:11 PM, Ruediger Pluem wrote:
> >>
> >>
> >> On 7/5/24 2:04 PM, Stefan Eissing via dev wrote:
> >>>
> >>>
> >>>> Am 05.07.2024 um 13:51 schrieb Ruediger Pluem <rpl...@apache.org>:
> >>>>
> >>>> I just noticed that mod_md in 2.4.61 fails to compile with openssl < 
> >>>> 1.1.1. Below is the output against openssl 1.0.2 on RedHat 7:
> >>>>
> >>>> md_crypt.c: In function 'md_pkey_get_rsa_e64':
> >>>> md_crypt.c:982:5: warning: implicit declaration of function 
> >>>> 'EVP_PKEY_get0_RSA' [-Wimplicit-function-declaration]
> >>>>     const RSA *rsa = EVP_PKEY_get0_RSA(pkey->pkey);
> >>>>     ^
> >>>> md_crypt.c:982:22: warning: initialization makes pointer from integer 
> >>>> without a cast [enabled by default]
> >>>>     const RSA *rsa = EVP_PKEY_get0_RSA(pkey->pkey);
> >>>>                      ^
> >>>> md_crypt.c: In function 'md_pkey_get_rsa_n64':
> >>>> md_crypt.c:1002:22: warning: initialization makes pointer from integer 
> >>>> without a cast [enabled by default]
> >>>>     const RSA *rsa = EVP_PKEY_get0_RSA(pkey->pkey);
> >>>>                      ^
>
> This was already the case with 2.4.59 and openssl 1.0.2. Hence we did not 
> fail to compile but loading of mod_md likely would fail
> as the symbol EVP_PKEY_get0_RSA is not available with openssl 1.0.2.

This probably comes from r1913912 (2.4.x) which backported r1913616
(trunk) which changed EVP_PKEY_get1_RSA() => EVP_PKEY_get0_RSA(), the
former being probably available in < 1.1.1.
So the check for using EVP_PKEY_get{0,1}_RSA() or the new openssl >= 3
API should probably be something like:

#if OPENSSL_VERSION_NUMBER < 0x10101000L
    RSA *rsa = EVP_PKEY_get1_RSA(pkey->pkey);
    if (rsa) {
        const char *ret;
        const BIGNUM *e;
        RSA_get0_key(rsa, NULL, &e, NULL);
        ret = bn64(e, p);
        RSA_free(rsa);
        return ret;
    }
#elif OPENSSL_VERSION_NUMBER < 0x30000000L
    ...
#else
    ...
#endif

?


Regards;
Yann.

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