Section 1.3.1

[dibyendumajumdar]

Hi,

The book says 'The writes into 32-bit parts, however, fill the upper 32 bits of the full register with sign bits. For example, zeroing eax will zero the entire rax, storing -1 into eax will fill the upper 32 bits with ones.'.

However Intel manual says in section 3.4.1.1:

32-bit operands generate a 32-bit result, zero-extended to a 64-bit result in the destination general-purpose register

I checked this in Visual Studio using a simple program:

.code
SomeFunction proc
	mov eax, -1
	ret
SomeFunction endp
end

The RAX register was zero extended as described in Intel manual.

Is this an error in the book?

Thanks and Regards
Dibyendu

[sayon]

Hello,
it is indeed an error. A similar error for another section is already present in errata, but not this specific one.
I will add it to the errata.

Thank you very much for reporting and helping to get rid of errors for a better experience!

[ghost]

An experiment in my x64 computer.

mov rax,0
   mov al,+1	rax = 0000000000000001
   mov ah,+1	rax = 0000000000000100
   mov ax,+1	rax = 0000000000000001
   mov eax,+1	rax = 0000000000000001
   mov rax,+1	rax = 0000000000000001

mov rax,0
   mov al,-1	rax = 00000000000000FF
   mov ah,-1	rax = 000000000000FF00
   mov ax,-1	rax = 000000000000FFFF
   mov eax,-1	rax = 00000000FFFFFFFF
   mov rax,-1	rax = FFFFFFFFFFFFFFFF

mov rax,-1
   mov al,+1	rax = FFFFFFFFFFFFFF01
   mov ah,+1	rax = FFFFFFFFFFFF01FF
   mov ax,+1	rax = FFFFFFFFFFFF0001
   mov eax,+1	rax = 0000000000000001
   mov rax,+1	rax = 0000000000000001

mov rax,-1
   mov al,-1	rax = FFFFFFFFFFFFFFFF
   mov ah,-1	rax = FFFFFFFFFFFFFFFF
   mov ax,-1	rax = FFFFFFFFFFFFFFFF
   mov eax,-1	rax = 00000000FFFFFFFF
   mov rax,-1	rax = FFFFFFFFFFFFFFFF