3u基本完美模拟了但不想挖矿。。。。
打开ce找个程序打开然后在'数值类型'上右键添加新类型[自动汇编],然后将下面的代码复制粘贴:
alloc(TypeName,256)
alloc(ByteSize,4)
alloc(ConvertRoutine,1024)
alloc(ConvertBackRoutine,1024)
TypeName:
db '4 Byte Big Endian',0
ByteSize:
dd 4
//The convert routine should hold a routine that converts the data to an integer (in eax)
//function declared as: stdcall int ConvertRoutine(unsigned char *input);
//Note: Keep in mind that this routine can be called by multiple threads at the same time.
ConvertRoutine:
//jmp dllname.functionname
[64-bit]
//or manual:
//parameters: (64-bit)
//rcx=address of input
xor eax,eax
mov eax,[rcx] //eax now contains the bytes 'input' pointed to
bswap eax //convert to big endian
ret
[/64-bit]
[32-bit]
//jmp dllname.functionname
//or manual:
//parameters: (32-bit)
push ebp
mov ebp,esp
//[ebp+8]=input
//example:
mov eax,[ebp+8] //place the address that contains the bytes into eax
mov eax,[eax] //place the bytes into eax so it's handled as a normal 4 byte value
bswap eax
pop ebp
ret 4
[/32-bit]
//The convert back routine should hold a routine that converts the given integer back to a row of bytes (e.g when the user wats to write a new value)
//function declared as: stdcall void ConvertBackRoutine(int i, unsigned char *output);
ConvertBackRoutine:
//jmp dllname.functionname
//or manual:
[64-bit]
//parameters: (64-bit)
//ecx=input
//rdx=address of output
//example:
bswap ecx //convert the little endian input into a big endian input
mov [rdx],ecx //place the integer the 4 bytes pointed to by rdx
ret
[/64-bit]
[32-bit]
//parameters: (32-bit)
push ebp
mov ebp,esp
//[ebp+8]=input
//[ebp+c]=address of output
//example:
push eax
push ebx
mov eax,[ebp+8] //load the value into eax
mov ebx,[ebp+c] //load the address into ebx
//convert the value to big endian
bswap eax
mov [ebx],eax //write the value into the address
pop ebx
pop eax
pop ebp
ret 8
[/32-bit]
打开cemu进入游戏再打开ce,把数值类型选为'4 byte big endian'也就是刚才添加的选项,然后搜索'07000001'十六进制,在搜索结果中选择'16xxx6c00'这项上右键选择浏览内存
按照图中标记修改即可,附上代码[天之护石种类填0A]
下图就是匠10利10的护石。。挖矿终结\(^ω^\)
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