movq and 64 bit numbers

Question

When I write to a register, everything is fine,

movq  $0xffffffffffffffff, %rax

But I get Error: operand size mismatch when I write to a memory location,

movq  $0xffffffffffffffff, -8(%rbp)

Why is that? I see in compiled C code that in asm these numbers are split in two and two movl instructions show up.

Maybe you can tell me where the mowq and other instructions are documented.

Answer 1

Why is that?

Because MOV r64, imm64 is a valid x86 instruction, but MOV r/m64, imm64 is not (there's no encoding for it).

I see in compiled C code that in asm these numbers are split in two and two movl instructions show up.

MOV r/m64, imm32 is a valid x86 instruction, which is why you see two of them being used to store a 64-bit immediate to memory.

Maybe you can tell me where the mowq and other instructions are documented

In Intel's Software Developer Manuals.