1

使用此代码，我猜扫描速度更快，但扫描始终返回相同的地址。

例如：

00123456
00124567
00135478
00145893
00123456 //start repeat 
00124567
00135478
00145893
00123456 //start repeat 
00124567
00135478
00145893

这是我的程序：

procedure SCANBYTE(value: integer);
var
 lpflOldProtect: dword;
 s: size_t;
 mbi: MEMORY_BASIC_INFORMATION;
 SI: SYSTEM_INFO;
 lpStartAddress, lpStopAddress: dword;
 addr: dword;
 i: dword;
begin
 GetSystemInfo(si);
 lpStartAddress := dword(SI.lpMinimumApplicationAddress);
 lpStopAddress := dword(SI.lpMaximumApplicationAddress);
 for addr := lpStartAddress to lpStopAddress do begin
  S:= VirtualQuery(Pointer(addr), MBI, SizeOf(MEMORY_BASIC_INFORMATION));
  if (S=SizeOf(MEMORY_BASIC_INFORMATION)) and (MBI.State = MEM_COMMIT) and (MBI.Type_9 = MEM_PRIVATE) and (MBI.RegionSize>0) and (MBI.Protect = PAGE_READWRITE) then begin
   for i := dword(MBI.BaseAddress) to (dword(MBI.BaseAddress) + dword(MBI.RegionSize)) - 4096 do begin
     if value = PBYTE(i)^ then ListBox1.Items.Add(IntToHex(i,8));
   end;
  end;
 end;
end;

我猜问题出在最后一个 FOR 循环：

(...)
for i := dword(MBI.BaseAddress) to (dword(MBI.BaseAddress) + dword(MBI.RegionSize)) - 4096 do begin
(...)

但我真的不知道..我该如何解决这个问题？

4

1 回答 1

8

您在从起始地址到结束地址的循环中运行代码。每次循环时，地址都会addr增加1 。VirtualQuery为您提供有关整个页面的信息。页中的所有地址都具有相同的基地址。文档告诉您，“此值向下舍入到下一页边界。”

Look more closely, and you should see that mbi.BaseAddress remains the same for 4096 iterations of your outer loop (assuming 4096 is the page size). Thus, you're re-scanning the same block of memory over and over again. (That might also explain why your code is slow.)

于 2012-04-04T19:43:48.027 回答