cryptography for developers 2006 phần 2 pdf

cryptography for developers 2006 phần 2 pdf

cryptography for developers 2006 phần 2 pdf

... Chapter 2 • ASN.1 Encoding 404_CRYPTO_ 02. qxd 10 /27 /06 3:40 PM Page 44 Table 2. 4 Example INTEGER Encodings Value Encoding 0 0x 02 01 00 1 0x 02 01 01 2 0x 02 01 02 127 0x 02 01 7F 128 0x 02 02 00 80 –1 ... { 019 return -2; 020 } 021 022 /* get # of bits */ 023 nbits = ((payload_length - 1) << 3) - in[0]; 024 ++in; 025 026 /* too many? */ 027 if (nbits > *o...

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cryptography for developers 2006 phần 1 potx

cryptography for developers 2006 phần 1 potx

... companies. KEY SERIAL NUMBER 001 HJIRTCV764 0 02 PO9873D5FG 003 829 KM8NJH2 004 GPPQQW 722 M 005 CVPLQ6WQ23 006 VBP965T5T5 007 HJJJ863WD3E 008 29 87GVTWMK 009 629 MP5SDJT 010 IMWQ295T6T PUBLISHED BY Syngress Publishing, ... . . . . . . . . . .22 5 SHA-5 12 State . . . . . . . . . . . . . . . . . . . . . . . . . . . .22 6 SHA-5 12 Expansion . . . . . . . . . . . . . . . . . . . . . . . .22...

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cryptography for developers 2006 phần 3 potx

cryptography for developers 2006 phần 3 potx

... 1. Lag[ 1] = 524 999 (expected 524 287) Lag[ 2] = 525 284 (expected 524 287) Lag[ 3] = 524 638 (expected 524 286) Lag[ 4] = 525 564 (expected 524 286) Lag[ 5] = 524 480 (expected 524 285) Lag[ 6] = 523 917 (expected ... test. Lag[ 1] = 26 2638 (expected 524 287) Lag[ 2] = 393784 (expected 524 287) Lag[ 3] = 459840 (expected 524 286) Lag[ 4] = 4 921 75 (expected 524 286) Lag[ 5] = 50 823 2...

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cryptography for developers 2006 phần 4 docx

cryptography for developers 2006 phần 4 docx

... a a a 0,0 0,1 0 ,2 0,3 1,0 1,1 1 ,2 1,3 2, 0 2, 1 2, 2 2, 3 3,0 3,1 3 ,2 3,3 b b b b b b b b b b b b b b b b 0,0 0,1 0 ,2 0,3 1,0 1,1 1 ,2 1,3 2, 0 2, 1 2, 2 2, 3 3,0 3,1 3 ,2 3,3 ROL 1 ROL 2 ROL 3 a a a a a ... a a a a a a a 0,0 0,1 0 ,2 0,3 1,0 1,1 1 ,2 1,3 2, 0 2, 1 2, 2 2, 3 3,0 3,1 3 ,2 3,3 b b b b b b b b b b b b b b b b 0,0 0,1 0 ,2 0,3 1,0 1,1 1 ,2 1,3 2, 0...

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cryptography for developers 2006 phần 5 ppt

cryptography for developers 2006 phần 5 ppt

... y++) printf("%02x ", blk[y]); 22 0 printf("\n"); 22 1 return -1; 22 2 } 22 3 } 22 4 } 22 5 printf("AES passed\n"); 22 6 return 0; 22 7 } This implementation will serve as ... 9: 42 AM Page 176 20 5 (Te4 _2[ byte(t3, 2) ]) ^ 20 6 (Te4_1[byte(t0, 1)]) ^ 20 7 (Te4_0[byte(t1, 0)]) ^ 20 8 rk [2] ; 20 9 STORE32H(s2, ct+8); 21 0 s3 = 21 1 (Te4_3[byte(t3...

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cryptography for developers 2006 phần 6 potx

cryptography for developers 2006 phần 6 potx

... *dst) 21 7 { 21 8 sha5 12_ state md; 21 9 sha5 12_ init(&md); 22 0 sha5 12_ process(&md, in, len); 22 1 sha5 12_ done(&md, dst); 22 2 } This is our familiar helper function to perform a SHA-5 12 compression ... buffer. 22 4 #include <stdio.h> 22 5 #include <stdlib.h> 22 6 #include <string.h> 22 7 int main(void) 22 8 { 22 9 static const struct { 23 0...

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cryptography for developers 2006 phần 7 doc

cryptography for developers 2006 phần 7 doc

... 64) { 018 sha1_memory(key, keylen, K); 019 i = 20 ; 020 } else { 021 /* copy key */ 022 for (i = 0; i < keylen; i++) { 023 K[i] = key[i]; 024 } 025 } If the secret key is larger than 64 bytes ... 0x0c}, 20 , 117 “Test With Truncation”, 20 , 118 {0x4c, 0x1a, 0x03, 0x 42, 0x4b, 0x55, 0xe0, 0x7f, 119 0xe7, 0xf2, 0x7b, 0xe1, 0xd5, 0x8b, 0xb9, 0x 32, 120 0x4a, 0x9a, 0x5a, 0x04} },...

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cryptography for developers 2006 phần 8 ppsx

cryptography for developers 2006 phần 8 ppsx

... 11:51 AM Page 317 21 5 STORE32H(tmp[i], pTmp + (i<< ;2) ); 21 6 } 21 7 #else 21 8 for (i = 0; i < 4; i++) { 21 9 STORE64H(tmp[i], pTmp + (i<<3)); 22 0 } 22 1 #endif 22 2 22 3 /* reduce by taking ... CRYPT_INVALID_ARG; 020 } 021 022 if ((err = cipher_is_valid(gcm->cipher)) != CRYPT_OK) { 023 return err; 024 } 025 026 027 if (gcm->mode != GCM_MODE_TEXT)...

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cryptography for developers 2006 phần 9 potx

cryptography for developers 2006 phần 9 potx

... the encryption. 21 2 x = 0; 21 3 CTRlen = 16; 21 4 21 5 /* now handle the PT */ 21 6 if (ptlen > 0) { 21 7 y = 0; 21 8 #ifdef LTC_FAST 21 9 if (ptlen & ~15) { 22 0 if (direction == CCM_ENCRYPT) { 22 1 for (; ... 0; 199 20 0 /* flags */ 20 1 ctr[x++] = L-1; 20 2 20 3 /* nonce */ 20 4 for (y = 0; y < (16 - (L+1)); ++y) { 20 5 ctr[x++] = nonce[y]; 20 6 } 20 7 /* of...

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cryptography for developers 2006 phần 10 doc

cryptography for developers 2006 phần 10 doc

... codes) advantage, 25 4, 25 7, 28 3, 29 4 authentication, 28 2 29 3 birthday attacks, 25 3 description, 9, 24 0 24 1, 25 2, 29 3 hash functions, 27 8 key lifespan, 25 4 patents, 29 6 purpose, 25 2 25 3 RNG processing, 27 8 security ... algorithms, 27 9 consequences, 27 6 27 8 counters, 28 0 28 1 description, 9–10, 23 6 23 8 design, 26 8 27 0 encryption, 28 1 29 2 history,...

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