What is the meaning of 'mov %reg8, (%reg32)' (not movb)?

Question

the test platform is x86-32bit.

I know that on x86-32bit, we have two opcodes movsbl and movzbl which has the following semantics:

%eax = 0x12345678
%edx = 0xAAAABBBB
MOVB %dh, %al         %eax = 0x123456BB
MOVSBL %dh, %eax      %eax = 0xFFFFFFBB
MOVZBL %dh, %eax      %eax = 0x000000BB

The above example is from here.

Then I am kind of confused with the semantics of the following instruction:

mov %dl, 0x2c(%esp)

What is the exact meaning of the above mov, is it equal to movsbl ? or equal to movzbl? Or neither?

Neither, that is a simple 8 bit move. It will only write a single byte of memory. Note it is not `mem32` it's `mem8` (implicit due to the 8 bit register used). — Jester, Sep 14 '15 at 19:03
Thank you @Jester. So you mean it actually equals to `movb %dl, 0x2c(%esp)` ? — lllllllllllll, Sep 14 '15 at 19:10
Yes, it generates the exact same machine code. The assembler deduced the `b` size from the register used. — Jester, Sep 14 '15 at 19:12

owacoder · Accepted Answer · 2015-09-15T03:46:09.207

If no operand sizes are explicitly provided, most assemblers will calculate the type of operation from the operand sizes. Therefore, in this case, mov %dl, 0x2c(%esp) is equivalent to movb %dl, 0x2c(%esp), a simple 1 byte move, deducing the b suffix from the one-byte register, dl.

The reason for the 32 bit register: this stores the address of the memory location; no mismatch in operand sizes results (since a memory location can be interpreted as being any size).

What is the meaning of 'mov %reg8, (%reg32)' (not movb)?

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