AES解密算法的问题……

我猜你應該是在 modulus 上面有問題
今天假設 A* B  的結果是C
而你還要針對 C 去做 modulus

module 都能写错……

没啥说的啦  飘过 ……都麻木了

http://www.eetop.cn/bbs/thread-239812-1-1.html
有AES算法源码。

小编，有需求可以讨论，本人有刚写好的，比较实用，比网上疯传的那个好。哈哈
QQ：809174805

1# 1920
去中国期刊网下载相关硕士论文，所有问题都有的。

AES是个比较成熟的算法，去开源上看看估计应该是有的。

逆列混合其实就是有限域乘法问题，而且是变量与固定数乘法，分析一下有限域乘法怎么做，自然就知道invmixcol怎么实现了。
Finite Field Multiplication
An example is shown as below.
{57}•{83} = {57}•{01} ^ {57}•{02} ^ {57}•{80}
{57}•{01} = {57}
{57}•{02} = {AE} //(Shift 1 bit to left, and XOR{1b} if the MSB before shifting is 1)
{57}•{04} = {AE}•{02} = {1}{01011100}
{57}•{08} = {47}•{02} = {8E}
{57}•{10} = {8E }•{02} = {1}{00011100}
{57}•{20} ={07}•{02} = {0E}
{57}•{40} ={0E}•{02} = {1C}
{57}•{80} ={1C }•{02} = {38}
{57}•{83} = {57} ^ {AE} ^{38} = {C1}
Another way to calculate {57}•{83} which is suitable for hardware implementation.
{57}•{83}
= {57}•{01}^ {57}•{02} ^ {57}•{80}
= {01010111} ^ {10101110} ^ {0101011}{10000000}
= {0101011}{01111001}
Step1: {0101011}{01111001}
= {01011}{01111001} ^ {00011}{01100000}
= {01000}{00011001}
Step2: {01000}{00011001}
= {000}{00011001}^{000}{11011000}
= {11000001} = {C1}

逆列混合其实就是有限域乘法问题，而且是变量与固定数乘法，分析一下有限域乘法怎么做，自然就知道invmixcol怎么实现了。
Finite Field Multiplication
An example is shown as below.
{57}•{83} = {57}•{01} ^ {57}•{02} ^ {57}•{80}
{57}•{01} = {57}
{57}•{02} = {AE} //(Shift 1 bit to left, and XOR{1b} if the MSB before shifting is 1)
{57}•{04} = {AE}•{02} = {1}{01011100}
{57}•{08} = {47}•{02} = {8E}
{57}•{10} = {8E }•{02} = {1}{00011100}
{57}•{20} ={07}•{02} = {0E}
{57}•{40} ={0E}•{02} = {1C}
{57}•{80} ={1C }•{02} = {38}
{57}•{83} = {57} ^ {AE} ^{38} = {C1}
Another way to calculate {57}•{83} which is suitable for hardware implementation.
{57}•{83}
= {57}•{01}^ {57}•{02} ^ {57}•{80}
= {01010111} ^ {10101110} ^ {0101011}{10000000}
= {0101011}{01111001}
Step1: {0101011}{01111001}
= {01011}{01111001} ^ {00011}{01100000}
= {01000}{00011001}
Step2: {01000}{00011001}
= {000}{00011001}^{000}{11011000}
= {11000001} = {C1}

逆列混合其实就是有限域乘法问题，而且是变量与固定数乘法，分析一下有限域乘法怎么做，自然就知道invmixcol怎么实现了。
Finite Field Multiplication
An example is shown as below.
{57}•{83} = {57}•{01} ^ {57}•{02} ^ {57}•{80}
{57}•{01} = {57}
{57}•{02} = {AE} //(Shift 1 bit to left, and XOR{1b} if the MSB before shifting is 1)
{57}•{04} = {AE}•{02} = {1}{01011100}
{57}•{08} = {47}•{02} = {8E}
{57}•{10} = {8E }•{02} = {1}{00011100}
{57}•{20} ={07}•{02} = {0E}
{57}•{40} ={0E}•{02} = {1C}
{57}•{80} ={1C }•{02} = {38}
{57}•{83} = {57} ^ {AE} ^{38} = {C1}
Another way to calculate {57}•{83} which is suitable for hardware implementation.
{57}•{83}
= {57}•{01}^ {57}•{02} ^ {57}•{80}
= {01010111} ^ {10101110} ^ {0101011}{10000000}
= {0101011}{01111001}
Step1: {0101011}{01111001}
= {01011}{01111001} ^ {00011}{01100000}
= {01000}{00011001}
Step2: {01000}{00011001}
= {000}{00011001}^{000}{11011000}
= {11000001} = {C1}

不好意思，还以为没发出去，一下子发了这么多。
逆行移位就更简单了，直接以为就是了
row[31:0] = {row[7:0, row[31:8]};
这就是移位

aes不都是有现成的实现库吗?如crypt++，

乘法的种类就那么几种，就是有限域乘法，先用xtime，然后xor