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  • 大小: 4KB
    文件類型: .rar
    金幣: 2
    下載: 0 次
    發布日期: 2021-05-28
  • 語言: Matlab
  • 標簽: matlab??AES??

資源簡介

csdn找了半天沒有直接可以用的代碼。這個壓縮包解開放一個文件夾 可以直接用 。直接在command 里面輸入[s_box, inv_s_box, w, poly_mat, inv_poly_mat] = aes_init; 2 plaintext_hex = {’00’ ’11’ ’22’ ’33’ ’44’ ’55’ ’66’ ’77’ ... 3 ’88’ ’99’ ’aa’ ’bb’ ’cc’ ’dd’ ’ee’ ’ff’}; 4 plaintext = hex2dec (plaintext_hex); 5 ciphertext = cipher (plaintext, w, s_box, poly_mat); 6 re_plaintext = inv_cipher (ciphertext, w, inv_s_box, inv_poly_mat); plaintext 明文 cipehertext密文 re_plaintext解密 key在 aes_init修改 實測可用

資源截圖

代碼片段和文件信息

function?state_out?=?add_round_key?(state_in?round_key)
state_out?=?bitxor?(state_in?round_key);

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件?????????99??2017-12-21?14:21??Aes代碼\add_round_key.m

?????文件????????329??2017-12-21?14:36??Aes代碼\aes_init.m

?????文件????????212??2017-12-20?22:32??Aes代碼\aff_trans.m

?????文件????????578??2017-12-20?22:56??Aes代碼\cipher.m

?????文件????????293??2017-12-20?22:48??Aes代碼\cycle.m

?????文件????????166??2017-12-20?22:26??Aes代碼\find_inverse.m

?????文件????????613??2017-12-21?14:54??Aes代碼\inv_cipher.m

?????文件?????????86??2017-12-21?14:28??Aes代碼\inv_shift_rows.m

?????文件????????337??2017-12-20?22:42??Aes代碼\key_expansion.m

?????文件????????507??2017-12-21?14:25??Aes代碼\mix_columns.m

?????文件????????324??2017-12-20?22:54??Aes代碼\poly_mat_gen.asv

?????文件????????324??2017-12-20?22:56??Aes代碼\poly_mat_gen.m

?????文件????????351??2017-12-20?22:25??Aes代碼\poly_mult.m

?????文件????????176??2017-12-20?22:40??Aes代碼\rcon_gen.m

?????文件?????????58??2017-12-20?22:43??Aes代碼\rot_word.m

?????文件?????????81??2017-12-21?14:22??Aes代碼\shift_rows.m

?????文件?????????82??2017-12-20?22:43??Aes代碼\sub_bytes.m

?????文件????????261??2017-12-20?22:34??Aes代碼\s_box_gen.m

?????文件????????104??2017-12-20?22:33??Aes代碼\s_box_inversion.m

?????目錄??????????0??2017-12-21?14:34??Aes代碼

-----------?---------??----------?-----??----

?????????????????4981????????????????????20


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