- 08 12月, 2002 1 次提交
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由 Richard Levitte 提交于
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- 07 12月, 2002 1 次提交
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由 Andy Polyakov 提交于
Submitted by: <appro> Reviewed by: PR:
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- 06 12月, 2002 7 次提交
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由 Richard Levitte 提交于
anything.
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由 Richard Levitte 提交于
compatible, don't use ==, use = instead...
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
a first alternative to try
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由 Richard Levitte 提交于
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- 05 12月, 2002 14 次提交
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由 Andy Polyakov 提交于
Submitted by: Reviewed by: PR:
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
PR: 380
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由 Richard Levitte 提交于
PR: 376
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由 Richard Levitte 提交于
well to be able to use the CHIL engine. PR: 281
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
PR: 381
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由 Richard Levitte 提交于
rule for SHA stuff. PR: 381
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由 Richard Levitte 提交于
PR: 379
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由 Richard Levitte 提交于
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由 Bodo Möller 提交于
(length of the order of the group, not length of the actual key, which will be shorter in some cases). Submitted by: Nils Larsch
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由 Bodo Möller 提交于
Include X9.62 signature examples. Submitted by: Nils Larsch
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- 04 12月, 2002 9 次提交
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由 Lutz Jänicke 提交于
Submitted by: Christian Hohnstaedt <chohnstaedt@innominate.com> Reviewed by: PR:
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由 Richard Levitte 提交于
PR: 347
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由 Richard Levitte 提交于
PR: 348
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由 Richard Levitte 提交于
PR: 375
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由 Dr. Stephen Henson 提交于
Make ASN1_TYPE_get() work for V_ASN1_NULL type.
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由 Dr. Stephen Henson 提交于
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由 Dr. Stephen Henson 提交于
return the requested number of bytes when reading content.
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由 Richard Levitte 提交于
exit() in whatever way works for the intended platform, and define OPENSSL_EXIT() to have the old meaning (the name is of course because it's only used in the openssl program)
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由 Richard Levitte 提交于
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- 03 12月, 2002 3 次提交
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
gets defined). PR: 371
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由 Richard Levitte 提交于
the divisor was a bit more complex than I first saw. The lost bit can't just be discarded, as there are cases where it is important. For example, look at dividing 320000 with 80000 vs. 80001 (all decimals), the difference is crucial. The trick here is to check if that lost bit was 1, and in that case, do the following: 1. subtract the quotient from the remainder 2. as long as the remainder is negative, add the divisor (the whole divisor, not the shofted down copy) to it, and decrease the quotient by one. There's probably a nice mathematical proof for this already, but I won't bother with that, unless someone requests it from me.
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- 02 12月, 2002 2 次提交
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
proper implementation of bn_div_words() for VAX. If the tests go through well, the next step will be to test on Alpha.
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- 01 12月, 2002 3 次提交
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由 Richard Levitte 提交于
PR: 366
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由 Richard Levitte 提交于
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由 Richard Levitte 提交于
PR: 366
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