Question

mov [ecx+eax*4+000005ac],edx 怎么处理

其 mov [ecx+eax*4+000005ac],#9999 无效我想我可能搞错了 请问解决方法

Answer 1

可以先假设为0 找到基址后 可以在指针上+4 +8这样... 用遍历工具看是最好的了

Answer 2

eax是不是00000000？ 如果是的话就是ecx+5ac，然后就好弄了

Answer 3

数组.... 又见数组.....

Answer 4

[ENABLE]//code from here to '[DISABLE]' will be used to enable the cheatalloc(newmem,2048) //2kb should be enoughlabel(returnhere)label(originalcode)label(exit)00402853:jmp newmemnopnopreturnhere:newmem: //this is allocated memory, you have read,write,execute access//place your code hereoriginalcode:mov [ecx+eax*4+000005ac],edxexit:jmp returnhere[DISABLE]//code from here till the end of the code will be used to disable the cheatdealloc(newmem)00402853:mov [ecx+eax*4+000005ac],edx//Alt: db 89 94 81 AC 05 00 00