VirtualAlloc address

Question

I worte the code below

 void * ptr1 = VirtualAlloc((void*)0x70000000, 32*1024*1024, MEM_RESERVE, PAGE_READWRITE);
 void * ptr2 = VirtualAlloc((void*)0x80000000, 4*1024*1024, MEM_RESERVE, PAGE_READWRITE);

But VirtualAlloc fails and the value of ptr1,ptr2 are always NULL. I want to use the adress 0x70000000,0x80000000. Does anyone know how I can use these adress ?

score 5 · Accepted Answer · answered Apr 13 '12 at 22:28

5

This part of the memory is not accessible for usermode applications. From the following MSDN page:

User applications cannot call VirtualAlloc with shared heap address range (0x70000000 to 0x7fffffff) because this is read-only for user applications and read/write for kernel mode.

answered Apr 13 '12 at 22:28

Mike Kwan

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Your link is Windows CE. I don't believe it's true for Windows in general. See http://msdn.microsoft.com/en-us/library/windows/desktop/aa366887(v=vs.85).aspx – Mark Ransom Apr 13 '12 at 22:57
I seem to recall above 0x70000000 being reserved for the kernel on Windows but couldn't find a reference for this – Mike Kwan Apr 13 '12 at 23:44

score 0 · Answer 2 · answered Apr 13 '12 at 22:24

Are you sure some of the pages in the requested memory block are not already reserved (or committed)? VirtualAlloc cannot reserve a reserved page.

Why are 0x70000000 and 0x80000000 special? In the flat memory model, there is no reason why one address would be preferable to any other.

VirtualAlloc address

2 Answers2