cryptography for developers 2006 phần 4 docx

cryptography for developers 2006 phần 4 docx

cryptography for developers 2006 phần 4 docx

... Standard 14 11 13 9 a 9 14 11 13 a 13 9 14 11 a 11 13 9 14 a 0 1 2 3 40 4_CRYPTO_ 04. qxd 10/30/06 9 :42 AM Page 1 54 Figure 4. 4 AES SubBytes Function Figure 4. 5 AES Affine Transformation There are several ... read? */ 140 if (pool_len > 0) { 141 /* copy upto pool_len bytes */ 142 for (y = 0; y < pool_len && y < len; y++) { 143 *out++ = pool[pool_idx++]; 144...

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cryptography for developers 2006 phần 1 potx

cryptography for developers 2006 phần 1 potx

... 2:33 PM Page iv 40 4_CRYPTO_01.qxd 10/27/06 4: 40 PM Page 20 Tom St Denis, Elliptic Semiconductor Inc. and Author of the LibTom Project Simon Johnson Cryptography Developers for 40 4_CRYPTO_FM.qxd ... done all the time without one thought toward cryptography. www.syngress.com 2 Chapter 1 • Introduction 40 4_CRYPTO_01.qxd 10/27/06 4: 40 PM Page 2 40 4_CRYPTO_FM.qxd 10/30/06 2:33...

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cryptography for developers 2006 phần 2 pdf

cryptography for developers 2006 phần 2 pdf

... read_unsigned(unsigned char *dst, unsigned long len) 040 { 041 long tmp; 042 043 tmp = 0; 044 while (len ) { 045 tmp = (tmp << 8) | *dst++; 046 } 047 return tmp; 048 } 049 This function reads bytes and stores ... swap- ping from 0 to x–1, not 0 to x. 041 /* move pointer */ 042 out += t; 043 } else { 044 *out++ = 0x00; 045 } 046 if (y < inlen - 1) { 047 wordbuf = in...

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cryptography for developers 2006 phần 3 potx

cryptography for developers 2006 phần 3 potx

... & 1. Lag[ 1] = 5 249 99 (expected 5 242 87) Lag[ 2] = 5252 84 (expected 5 242 87) Lag[ 3] = 5 246 38 (expected 5 242 86) Lag[ 4] = 5255 64 (expected 5 242 86) Lag[ 5] = 5 244 80 (expected 5 242 85) Lag[ 6] = 523917 ... (expected 5 242 87) Lag[ 2] = 3937 84 (expected 5 242 87) Lag[ 3] = 45 9 840 (expected 5 242 86) Lag[ 4] = 49 2175 (expected 5 242 86) Lag[ 5] = 508232 (expected 5 2...

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cryptography for developers 2006 phần 5 ppt

cryptography for developers 2006 phần 5 ppt

... %rax 841 andl $42 949 67295, %eax 842 salq $8, %rcx 843 shrq $ 24, %rax 844 orq %rcx, %rax 845 andl $42 949 67295, %eax 846 xorq %rax, %r10 As we can see, GCC is doing a 32-bit rotation with a 64- bit data ... 0x76767676UL, <snip> 143 0xcecececeUL, 0x55555555UL, 0x28282828UL, 0xdfdfdfdfUL, 144 0x8c8c8c8cUL, 0xa1a1a1a1UL, 0x89898989UL, 0x0d0d0d0dUL, 145 0xbfbfbfbfUL, 0xe6e6e6e6U...

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cryptography for developers 2006 phần 6 potx

cryptography for developers 2006 phần 6 potx

... CONST 64( 0x113f9804bef90dae), CONST 64( 0x1b710b35131c471b), 088 CONST 64( 0x28db77f523 047 d 84) , CONST 64( 0x32caab7b40c7 249 3), 089 CONST 64( 0x3c9ebe0a15c9bebc), CONST 64( 0x431d67c49c100d4c), 090 CONST 64( 0x4cc5d4becb3e42b6), ... CONST 64( 0x1e376c085 141 ab53), 077 CONST 64( 0x2 748 774cdf8eeb99), CONST 64( 0x34b0bcb5e19b48a8), 078 CONST 64( 0x391c0cb3c5c95a63), CONST 64( 0x4ed8aa4a...

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cryptography for developers 2006 phần 7 doc

cryptography for developers 2006 phần 7 doc

... */ 037 for (i = 0; i < 16; i++) { 038 cmac->L[1][i] = cmac->L[0][i]; 039 } 040 m = cmac->L[1][0] & 0x80 ? 1 : 0; 041 042 /* shift */ 043 for (i = 0; i < 15; i++) { 044 cmac->L[1][i] ... We only need it for a short period of time, so there is no reason to copy it to the HMAC state. 042 /* start hash */ 043 sha1_init(&hmac->hash); 044 045 /* hash k...

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cryptography for developers 2006 phần 8 ppsx

cryptography for developers 2006 phần 8 ppsx

... sizeof(B[0])); 142 zeromem(B[M(1)], sizeof(B[M(1)])); 143 144 #ifdef ENDIAN_32BITWORD 145 for (i = 0; i < 4; i++) { 146 LOAD32H(B[M(1)][i], a + (i<<2)); 147 LOAD32L(pB[i], b + (i<<2)); 148 ... terminate the GHASH. 041 /* increment counter */ 042 for (y = 15; y >= 12; y ) { 043 if (++gcm->Y[y] & 255) { break; } 044 } 045 /* encrypt the counter */ 046...

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cryptography for developers 2006 phần 9 potx

cryptography for developers 2006 phần 9 potx

... if(z!=y&&z+y==x&&y<=z){ 041 ++f; 042 } 043 } 044 } 045 046 if (f <= 2) { 047 for (y = 0; y < N; y++) { 048 for (z = 0; z < N; z++) { 049 if (y<=z && (y+z)==x) { 050 ... long as cy is not zero. 041 } 042 043 /* now copy out */ 044 _c = c + pa; 045 tmpm = a->dp; 046 for (x = 0; x < pa+1; x++) { 047 *tmpm++ = *_c++; 048 } 049...

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cryptography for developers 2006 phần 10 doc

cryptography for developers 2006 phần 10 doc

... 238, 248 passwords, 238 patents, 249 performance considerations, 244 – 245 purpose, 238 random number generators (RNGs), 108, 238 re-applied, 243 standards, 249 unsalted passwords, 240 uses, 249 See ... ECC192 { 24, “ECC-192”, “FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFEFFFFFFFFFFFFFFFF”, “ 642 10519E59C80E70FA7E9AB72 243 049 FEB8DEECC 146 B9B1”, “FFFFFFFFFFFFFFFFFFFFFFFF99DEF836 146 BC9B1B4D2...

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