ARM NEON: How does VQABS determine overflow conditon

Question

Can somebody explain to me how the ARM NEON instruction VQABS operates. Per the documentation:

"VQABS returns the absolute value of each element in a vector. If any of the results overflow, they are saturated and the sticky QC flag (FPSCR bit[27]) is set."

If I apply this on a uint16x8 QWORD type then how does the processor determine that there is an overflow? I am puzzled because there is NO operation being performed other than finding the absolute value of a 16-bit data which surely cannot "overflow".

VQABS only operates on signed types - you already have the absolute value of an unsigned type, by definition. Now, put NEON to one side for a moment and just think basic twos complement; in only 16 bits, what's abs(-32768)? — Notlikethat, Jul 16 '15 at 19:33
Understood. My question is valid even for signed values: How does one determine overflow when there is no arithmetic operation being performed? — Rocky, Jul 17 '15 at 01:26
But there _is_ an arithmetic operation (negation) being performed if the value is negative, that's the first misconception here. As for the second, like I said, try calculating abs(-32768) as a signed 16-bit value and see what happens. — Notlikethat, Jul 17 '15 at 07:28

score 1 · Answer 1 · answered Jul 17 '15 at 14:41

1

8bit: -128 ~ +127

16bit: -32768 ~ +32767

....

As you can see, the biggest negative number is always bigger than the positive one by 1.

If you use vabs on -128 (0x80) for example, the return value is the same 0x80 which can be interpreted both way: -128 as well as +128, depending on the signness.

It might be fine as long as you interpret the result as unsigned which is mostly the case. However, 0x80 needs one more bit than 0x7f (127), thus overflow might occur in following arithmetics that expect a 7-bit input.

Or you could be simply forced to do multiply the result with a signed vector then you would be in serious trouble since 0x80 will be interpreted as -128 by a signed multiply.

vqabs returns 0x7f instead of 0x80. Problem solved.

answered Jul 17 '15 at 14:41

Jake 'Alquimista' LEE

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So what you are saying is that the overflow bit will be set IF and ONLY IF the value in the register operand is 0x80 (assuming 8-bit register)? – Rocky Jul 18 '15 at 18:38
And thanks for the clarification about 0x7F versus 0x80. Just what I was looking for. – Rocky Jul 18 '15 at 19:56
Yes, only in case of inputs 0x80, 0x8000, 0x80000000, depending on the element size (8, 16, 32) – Jake 'Alquimista' LEE Jul 21 '15 at 13:19

ARM NEON: How does VQABS determine overflow conditon

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