Question

I was looking through the disassmbly of my program (because it crashed), and noticed lots of

xchg    ax, ax

I googled it and found out it’s essentially a nop, but why does visual studio do an xchg instead of a noop?

The application is a C# .NET3.5 64-bit application, compiled by visual studio

Answer 1

On x86 the NOP instruction is XCHG AX, AX

The 2 mnemonic instructions assemble to the same binary op-code. (Actually, I suppose an assembler could use any xchg of a register with itself, but AX or EAX is what’s typically used for the nop as far as I know).

xchg ax, ax has the properties of changing no register values and changing no flags (hey – it’s a no op!).

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Edit (in response to a comment by Anon.):

Oh right – now I remember there are several encodings for the xchg instruction. Some take a mod/r/m set of bits (like many Intel x86 architecture instructions) that specify a source and destination. Those encodings take more than one byte. There’s also a special encoding that uses a single byte and exchanges a general purpose register with (E)AX. If the specified register is also (E)AX then you have a single-byte NOP instruction. you can also specify that (E)AX be exchanged with itself using the larger variant of the xchg instruction.

I’m guessing that MSVC uses the multiple byte version of xchg with (E)AX as the source and destination when it wants to chew up more than one byte for no operation – it takes the same number of cycles as the single byte xchg, but uses more space. In the disassembly you won’t see the multiple byte xchg decoded as a NOP, even if the result is the same.

Specifically xchg eax, eax or nop could be encoded as opcodes 0x90 or 0x87 0xc0 depending on whether you want it to use up 1 or 2 bytes. The Visual Studio disassembler (and probably others) will decode the opcode 0x90 as the NOP instruction and will decode opcode 0x87 0xc0 as xchg eax, eax.

It’s been a while since I’ve done detailed assembly language work, so chances are I’m wrong on at least one count here…