2022-08-28 09:35:46 +01:00
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/*
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2022-11-14 14:02:28 +00:00
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Copyright 2001-2022 John Wiseman G8BPQ
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2022-08-28 09:35:46 +01:00
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This file is part of LinBPQ/BPQ32.
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LinBPQ/BPQ32 is free software: you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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LinBPQ/BPQ32 is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with LinBPQ/BPQ32. If not, see http://www.gnu.org/licenses
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*/
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#define WIN32_LEAN_AND_MEAN // Exclude rarely-used stuff from Windows headers
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#define _CRT_SECURE_NO_DEPRECATE
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#ifdef LINBPQ
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#include "compatbits.h"
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2024-12-16 17:54:16 +00:00
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char * strlop(const char * buf, char delim);
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2022-08-28 09:35:46 +01:00
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#define APIENTRY
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#define VOID void
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#else
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#include <windows.h>
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#endif
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VOID APIENTRY md5 (char *arg, unsigned char * checksum);
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// Implementation of the WinLink password challenge/response protocol
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unsigned char seed [] = {77, 197, 101, 206, 190, 249,
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93, 200, 51, 243, 93, 237,
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71, 94, 239, 138, 68, 108,
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70, 185, 225, 137, 217, 16,
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51, 122, 193, 48, 194, 195,
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198, 175, 172, 169, 70, 84,
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61, 62, 104, 186, 114, 52,
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61, 168, 66, 129, 192, 208,
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187, 249, 232, 193, 41, 113,
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41, 45, 240, 16, 29, 228,
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208, 228, 61, 20, 0};
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/*
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Calculate the challenge password response as follows:
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- Concatenate the challenge phrase, password, and supplied secret value (i.e. the salt)
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- Generate an MD5 hash of the result
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- Convert the first 4 bytes of the hash to an integer (big endian) and return it
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*/
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int GetCMSHash(char * Challenge, char * Password)
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{
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unsigned char Hash[16];
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unsigned char Phrase[256];
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strlop(Challenge, 13);
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strlop(Password, 13);
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strcpy(Phrase, Challenge);
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strcat(Phrase, Password);
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strcat(Phrase, seed);
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md5(Phrase, Hash);
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return ((Hash[3] & 0x3f) << 24) + (Hash[2] << 16) + (Hash[1] << 8) + Hash[0];
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}
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