In function ASN1_STRING_to_UTF8 why do you check for NULL on |!*out|? Why
can't the character pointer point to NULL? If I understand your function
correctly it allocates the memory so NULL makes perfect sense. Maybe it
should be |!out| so you know that |out| actually points to something? My
testing is with 0.9.6d but a quick look at version 0.9.7 beta 2 looks like
it behaves the same way.
"crypto/asn1/a_strex.c" line 514
/* Utility function: convert any string type to UTF8, returns number of
bytes
* in output string or a negative error code
*/
int ASN1_STRING_to_UTF8(unsigned char **out, ASN1_STRING *in)
{
ASN1_STRING stmp, *str = &stmp;
int mbflag, type, ret;
if(!*out || !in) return -1;
EXAMPLES:
This code fails
unsigned char * in = NULL;
ASN1_STRING in = .....;
ASN1_STRING_to_UTF8(&out, &in);
but this works (copied from "crypto/pkcs12/p12_kiss.c" line 253)
if(fname) {
int len;
unsigned char *data;
len = ASN1_STRING_to_UTF8(&data, fname);
if(len > 0) {
X509_alias_set1(x509, data, len);
OPENSSL_free(data);
}
In addition, why can't it accept a string of type V_ASN1_UTF8STRING (value
12)? It always fails on
"crypto/asn1/a_mbstr.c" line 142 ASN1err(ASN1_F_ASN1_MBSTRING_COPY,
ASN1_R_UNKNOWN_FORMAT);
I realize all I have to do is copy the buffer manually as a workaround, but
I was trying to avoid treating it as a special case.
Thanks,
Christopher Nebergall
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