CODEWITHSUNDEEP
stringassemblyx86attendianness
Sunspawn
Sunspawn

Reputation: 837

Accessing and moving bytes in X86 assembly

I have several questions about memory and registers in X86 assembly:

  1. I have a string "abcdefgh", and register %eax holds a pointer to the string. Now I use movl (%eax), %edx to grab the first four bytes of the string into %edx. How are they stored in the register? Is the character d in the %dl register, or is it the character a?

  2. When using movb %eax, %dl for example, which of %eax's bytes does it actually move? The one in %al or the opposite one? Is it even possible to do this? Or should I use a pointer like this - movb (%eax), %dh - to take the first byte the pointer points to?

Upvotes: 1

Views: 8405

Answers (1)

m0skit0
m0skit0

Reputation: 25863

Assuming you're using the unusual GAS' syntax (source is the first operand, destination is the second one) and not Intel's :

How are they stored in the register? Is the character d in the %dl register, or is it the character a?

Since you're accessing the string as if it was a 32-bit number, endianness applies. x86 is little-endian so you get the least-significant byte at the lowest address, so DL will hold 'a' (0x61), and the whole EDX would be 0x64636261.

When using movb %eax, %dl for example, which of %eax's bytes does it actually move? The one in %al or the opposite one? Is it even possible to do this?

That would give an syntax error because operands are of different size. You can't move 32 bits to 8 bits.

Or should I use a pointer like this - movb (%eax), %dh - to take the first byte the pointer points to?

If you want to access the data pointed by EAX and not EAX itself, then yes, that should work.

Upvotes: 5

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