Converting from hex to LPCVOID cuts off half of the address

Question

I'm trying to read (and eventually write) memory of another process. I have the address (found using cheat engine) and I want to read it's value from my program, but I don't get the expected value. The address is 274A88A1630, but when I convert it to LPCVOID (which is required by ReadProcessMemory) I only get A88A1630 (which doesn't point to the memory I want)

I've tried converting using (LPCVOID) and (void*), both give the same result

int val = 0;
ReadProcessMemory(handle, (void*)0x274A88A1630, &val, sizeof(val), 0);
cout << val <<endl;

`LPCVOID` and `void*` are same. To fit that memory address you need a 64-bit pointer. — Asesh, May 04 '19 at 16:28
Apparently the address is a 64-bit value, and when you convert it to void*, which is 32 bit, the 274 loss. — Tony, May 04 '19 at 16:34
Are you on a 32-bit system? Or did you build your program as a 32-bit executable? — Some programmer dude, May 04 '19 at 16:41
Yes, I was compiling for 32-bit. Changed to 64 and it works. Thanks. — LoipesMas, May 04 '19 at 16:58
Related : [Is it possible to read process memory of a 64 bit process from a 32bit app?](https://stackoverflow.com/questions/5714297/) — Remy Lebeau, May 04 '19 at 17:42

score 1 · Accepted Answer · edited May 04 '19 at 20:04

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That happens because you have compiled your app as an x86 (32 bit/Win32) binary. Pointers (void* in this case) are 32 bits on x86. Pointers on x64 (64 bit/Win64) are 64 bits. 0x274A88A1630 is a 64 bit value, so you will therefore not encounter this problem if you compile your app for x64 (64 bit/Win64).

TLDR; A pointer value over 32 bits gets truncated to 32 bits when compiled for x86.

edited May 04 '19 at 20:04

IInspectable

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answered May 04 '19 at 16:34

bootloop

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Converting from hex to LPCVOID cuts off half of the address

1 Answers1