x86 Assembly: INC and DEC instruction and overflow flag

In x86 assembly, the overflow flag is set when an add or sub operation on a signed integer overflows, and the carry flag is set when an operation on an unsigned integer overflows.

However, when it comes to the inc and dec instructions, the situation seems to be somewhat different. According to this website, the inc instruction does not affect the carry flag at all.

But I can't find any information about how inc and dec affect the overflow flag, if at all.

Do inc or dec set the overflow flag when an integer overflow occurs? And is this behavior the same for both signed and unsigned integers?

============================= EDIT =============================

Okay, so essentially the consensus here is that INC and DEC should behave the same as ADD and SUB, in terms of setting flags, with the exception of the carry flag. This is also what it says in the Intel manual.

The problem is I can't actually reproduce this behavior in practice, when it comes to unsigned integers.

Consider the following assembly code (using GCC inline assembly to make it easier to print out results.)

int8_t ovf = 0;

__asm__
(
    "movb $-128, %%bh;"
    "decb %%bh;"
    "seto %b0;"
    : "=g"(ovf)
    :
    : "%bh"
);

printf("Overflow flag: %dn", ovf);

Here we decrement a signed 8-bit value of -128. Since -128 is the smallest possible value, an overflow is inevitable. As expected, this prints out: Overflow flag: 1

But when we do the same with an unsigned value, the behavior isn't as I expect:

int8_t ovf = 0;

__asm__
(
    "movb $255, %%bh;"
    "incb %%bh;"
    "seto %b0;"
    : "=g"(ovf)
    :
    : "%bh"
);

printf("Overflow flag: %dn", ovf);

Here I increment an unsigned 8-bit value of 255. Since 255 is the largest possible value, an overflow is inevitable. However, this prints out: Overflow flag: 0 .

Huh? Why didn't it set the overflow flag in this case?

---

The overflow flag is set when an operation would cause a sign change. Your code is very close. I was able to set the OF flag with the following (VC++) code:

char ovf = 0;

_asm {
    mov bh, 127
    inc bh
    seto ovf
}
cout << "ovf: " << int(ovf) << endl;

When BH is incremented the MSB changes from a 0 to a 1, causing the OF to be set.

This also sets the OF:

char ovf = 0;

_asm {
    mov bh, 128
    dec bh
    seto ovf
}
cout << "ovf: " << int(ovf) << endl;

Keep in mind that the processor does not distinguish between signed and unsigned numbers. When you use 2's complement arithmetic, you can have one set of instructions that handle both. If you want to test for unsigned overflow, you need to use the carry flag. Since INC/DEC don't affect the carry flag, you need to use ADD/SUB for that case.

---

Intel® 64 and IA-32 Architectures Software Developer's Manuals

Look at the appropriate manual Instruction Set Reference, AM. Every instruction is precisely documented.

Here is the INC section on affected flags:

The CF flag is not affected. The OF, SZ, ZF, AZ, and PF flags are set according to the result.

---

try changing your test to pass in the number rather than hard code it, then have a loop that tries all 256 numbers to find the one if any that affects the flag. Or have the asm perform the loop and exit out when it hits the flag and or when it wraps around to the number it started with (start with something other than 0x00, 0x7f, 0x80, or 0xFF).

EDIT

.globl inc
inc:
    mov $33, %eax

top:
    inc %al
    jo done
    jmp top

done:
    ret

.globl dec
dec:
    mov $33, %eax

topx:
    dec %al
    jo donex
    jmp topx

donex:
    ret

Inc overflows when it goes from 0x7F to 0x80. dec overflows when it goes from 0x80 to 0x7F, I suspect the problem is in the way you are using inline assembler.

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