What is the purpose of XORing a register with itself? [duplicate]

xor eax, eax will always set eax to zero, right? So, why does MSVC++ sometimes put it in my executable's code? Is it more efficient that mov eax, 0?

012B1002  in          al,dx 
012B1003  push        ecx  
    int i = 5;
012B1004  mov         dword ptr [i],5 
    return 0;
012B100B  xor         eax,eax 

Also, what does it mean to do in al, dx?

Yes, it is more efficient.

The opcode is shorter than mov eax, 0, only 2 bytes, and the processor recognizes the special case and treats it as a mov eax, 0 without a false read dependency on eax, so the execution time is the same.

Also to avoid 0s when compiled as used on shell codes for exploitation of buffer overflows, etc. Why avoid the 0 ? Well, 0 represents the end of string in c/c++ and the shell code would be truncated if the mean of exploitation is a string processing function or the like.

Btw im referring to the original question: "Any reason to do a “xor eax, eax”?" not what the MSVC++ compiler does.

Since there's some debate in the comments about how this is pertinent in the real world, see this article and this section on Wikipedia.

xor eax, eax is a faster way of setting eax to zero. This is happening because you're returning zero.

The in instruction is doing stuff with I/O ports. Basically reading a word of data from the port specified dx in and storing it in al. It's not clear why it is happening here. Here's a reference that seems to explain it in detail.