intel assembly TEST PF flag operation

Question

I'm was doing a manual operation with TEST (Parity flag operation), the problem it's that i can't get the right result, consider this:

ax = 256 = 0000 0001 0000 0000

so if i do:

test ah, 0x44

the PF flag operation should be:

0000 0000 = 0000 0001 & 0100 0100

PF = 0000 0000 XNOR 0000 0000

PF = 1111 1111?

I've followed the intel reference, according to this:

the question it's what i'm doing wrong?

Racquelracquet

If you are trying to set or reset PF flag you can do that with lahf instruction have a look at my answer #41336153 – Barney

@Doda No, I want to do the formula manually, the problem it's that i'm not getting the same result than the program operation... – Racquelracquet

You're reading the reference incorrectly. First, the parameter is TEMP[0:7] but you are using TEMP[15:8]. Second, BitwiseXNOR has only one parameter, so it's not clear why you are calculating 0000 0000 XNOR 0000 0000 (especially since the result has 8 bits, but PF is only one bit). What BitwiseXNOR means is to XOR all the bits together, and then negate the result. – Sensation

I don't understand Are you trying to set/reset PF flag ? – Barney

@Raymond Chen Can you post a example or link example please? i can't find a operation done like example, and you mean bit0 to bit7 xorged and then negated? something like this ~(0^0^0^0^0^0^0)? – Racquelracquet

@Doda No. just i want to do the operation manually... – Racquelracquet

Answer 1

BitwiseXNOR is a horizontal XNOR of the bits, producing a single bit. Remember that PF is only 1 bit wide (a flag in EFLAGS), so it makes no sense to write PF=1111 1111.

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It's the same parity calculation as always for instructions that "set flags according to the result", except that it's done on test's internal temporary result. (And as always, on the low 8 bits of it, regardless of operand size).

PF = 0 if the number of set bits is odd, PF = 1 if the number of set bits is even. So yes, the expression you posted in comments, ~(0^0^0^0^0^0^0^0) is correct.

See also:

• https://en.wikipedia.org/wiki/Parity_flag
• https://en.wikipedia.org/wiki/Parity_bit points out that parity is also the inverted (sum mod 2) of all the bits. (Because XOR is add-without-carry). Parity of an integer wider than 8 bits can be computed with popcnt eax, eax / not eax / and eax, 1. (Or use BMI andn eax, eax, ecx with ecx=1 set outside a loop to do the not-and part in one instruction.) Or just use the inverted parity value, where 1 = odd parity.

This answer has several examples of how PF is set from the horizontal-XOR of the bits in a result. In your case, the 8 bits are the AND result that test internally generates.

Scouring

yes, i had a bad logic about BitwiseXNOR(Temporary[0:7]); because that i'm was doing wrong... thank you... – Racquelracquet