为什么没有设置AF和SF？

Question

I'm reading Duntemann's book (3rd edition) and have just started learning x86 assembly. I'm using a variant of Fedora 23 (64 bit). The following is the code:

section .data
section .text
    global  _start
_start:
    nop
; Put your experiments between the two nops...
    mov eax,0FFFFFFFFh
    mov ebx,02Dh
    dec ebx
    inc eax 
; Put your experiments between the two nops...
    nop

My makefile is as follows:

sandbox: sandbox.o
    ld -o sandbox sandbox.o -melf_i386
sandbox.o: sandbox.asm
    nasm -f elf -g -F stabs sandbox.asm -l sandbox.lst

So you can see I've taken care to assemble a 32bit executable and not 64bit. The problem, however, is that prior to dec ebx instruction the AF and SF flag is not set contrary to what the book claimed. Running the program in insight shows me 32 bit registers which further assures the executable is 32bit. The following is status shown by gdb just before dec ebx instruction.

(gdb) info reg
eax            0xffffffff   -1
ecx            0x0  0
edx            0x0  0
ebx            0x2d 45
esp            0xffffce80   0xffffce80
ebp            0x0  0x0
esi            0x0  0
edi            0x0  0
eip            0x804806b    0x804806b <_start+11>
eflags         0x202    [ IF ]
cs             0x23 35
ss             0x2b 43
ds             0x2b 43
es             0x2b 43
fs             0x0  0
gs             0x0  0

Duntemann's version in Pg 217 shows that AF and SF flag is set. What is wrong with my code?

Answer 1

As Margaret Bloom commented, the mov instruction does not set the flags! In fact, it has absolutely no effect on the flags whatsoever, which makes it very useful in cases where you want to set the contents of a register or memory without clobbering the current state of the flags.

This is documented in the official Intel IA-32 ISA reference manual , but if you're like me, you're lazy. Fortunately, we are in luck. Several kind folks have uploaded portions of the official documentation to their websites, and these can be found easily by simply Googling "x86" plus the name of the instruction mnemonic. For example, my top result for "x86 mov" is this page . Note the following section:

Flags affected
None.

Other instructions will say different things there, of course. The add instruction sets virtually all of the flags, as you can see here .

So, looking at your code, execution will begin with the flags set to meaningless, garbage values. Technically, they will contain whatever values they were last set to, but this is meaningless in the context of your program, because you did not cause them to be set to anything, and therefore cannot rely on them being set to meaningful values!

Neither the nop nor the mov instructions will affect the values of the flags, so they will continue to contain garbage values.

The flags will not contain meaningful values until the dec instruction is executed. Then, as per the documentation , the overflow, sign, zero, adjust, and parity flags will be set according to the result of the decrement operation. The carry flag is not affected by dec and will therefore continue to contain a garbage value.

The inc instruction sets the flags in exactly the same way as dec , so the overflow, sign, zero, adjust, and parity flags will change, but the carry flag will not (and will continue to contain a garbage value).

---

I don't have a copy of the book you reference, so I don't know what it actually says. If it's just showing a dump of the register/flag state at that point in the code's execution, then the flags are just garbage values and not intended to be significant—you're supposed to focus on how the values in the registers have changed as a result of the mov instruction(s). If it is actually implying that the mov instruction sets the flags, well, then the book has a bug. :-)