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I think its correct and I don't expect it would change the only problem is that it might not be all of it, e.g there could be X amount more bytes needed to complete it.
As for camSegPtr
segDWord = *(_WORD *)(camSegPtr + 2) | (*(_WORD *)camSegPtr << 16);
v8 = camSegPtr + 4;
Yeah that looks like what it is, its switching the two words of the cam segment dword around and then advancing to the next cam segment dword (so that suggests the cam data is an array of either words or dwords).
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Yeah, that's why I asked if it's Little Endian. It would make sense if it's Little Endian, to read a double word instead of a word (already in LE).
Anyway, I've tried this function, and it doesn't work it seems (either a segfault, or an infinite loop, depending on the endianness I use).