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I am learning assembly and reading "Computer Systems: A programmer's perspective". In Practice Problem 3.3, it says movl %eax,%rdx will generate an error. The answer keys says movl %eax,%dx Destination operand incorrect size. I am not sure if this is a typo or not, but my question is: is movl %eax,%rdx a legal instruction? I think it is moving the 32 bits in %eax with zero extension to %rdx, which will not be generated as movzql since

an instruction generating a 4-byte value with a register as the destination will fill the upper 4 bytes with zeros` (from the book).

I tried to write some C code to generate it, but I always get movslq %eax, %rdx(GCC 4.8.5 -Og). I am completely confused.

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    both operands have to have the same size. dx is 16bit , eax 32 bits.
    – 0___________
    Commented Jul 2, 2022 at 19:18
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    The %dx in the answer is probably a typo and should be %rdx to match the question. Note that if you are using the Global Edition, its exercises and typos are filled with severe errors and should probably be ignored altogether. See stackoverflow.com/questions/57998998/…
    – Nate Eldredge
    Commented Jul 2, 2022 at 19:27
  • If you're getting movslq, apparently you used signed int instead of unsigned or uint32_t. C semantics require sign-extension when promoting a signed type to any wider type, preserving the value. So cast to unsigned first, or long foo(unsigned x){ return x; }. godbolt.org/z/YhevPfEvc
    – Peter Cordes
    Commented Jul 2, 2022 at 22:11

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The GNU assembler doesn't accept movl %eax,%rdx. It also doesn't make sense for the encoding, since mov must have a single operand size (using a prefix byte if needed), not two different sized operands.

The effect you want is achieved by movl %eax, %edx since writes to a 32-bit register always zero-extend into the corresponding 64-bit register. See Why do x86-64 instructions on 32-bit registers zero the upper part of the full 64-bit register?.

movzlq %eax, %rdx might make logical sense, but it's not supported since it would be redundant.

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  • Just to clarify, movl %eax, %edx is equivalent to movzlq %eax, %rdx so movzlq is redundant, right?
    – jeffma
    Commented Jul 2, 2022 at 19:43
  • @jeffma: I guess I would say that if movzlq %eax, %rdx did exist, it would be equivalent to movl %eax, %edx, which is why it does not exist.
    – Nate Eldredge
    Commented Jul 2, 2022 at 19:43
  • @jeffma: MOVZX missing 32 bit register to 64 bit register goes into more detail about why no special mnemonic for that exists, because 32-bit mov does the job.
    – Peter Cordes
    Commented Jul 2, 2022 at 22:08

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