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2ef3a5e)
This will solve the long running of the attack, since all candidate keys will be tested.
int size = keycount - i > max_keys ? max_keys : keycount - i;\r
for (int j = 0; j < size; j++) {\r
if (par_list == 0) {\r
int size = keycount - i > max_keys ? max_keys : keycount - i;\r
for (int j = 0; j < size; j++) {\r
if (par_list == 0) {\r
- num_to_bytes(last_keylist[i*max_keys + j], 6, keyBlock);\r
+ num_to_bytes(last_keylist[i*max_keys + j], 6, keyBlock+(j*6));\r
- num_to_bytes(keylist[i*max_keys + j], 6, keyBlock);\r
+ num_to_bytes(keylist[i*max_keys + j], 6, keyBlock+(j*6));\r
}\r
}\r
if (!mfCheckKeys(0, 0, false, size, keyBlock, key)) {\r
}\r
}\r
if (!mfCheckKeys(0, 0, false, size, keyBlock, key)) {\r