Sqrt ASM code with rcc 1.2.21

[jeandet]

Hello,
I’m trying to optimize some code for an instrument on Solar Orbiter. We have to do some square roots, so while reading ASM code to see some C code change effects and hunting for FPU instructions, I discovered that pattern for sqrt:

4001a9ac:	97 a0 05 2f 	fsqrts  %f15, %f11
4001a9b0:	81 aa ca 2b 	fcmps  %f11, %f11
4001a9b4:	01 00 00 00 	nop 
4001a9b8:	13 80 00 11 	fbe  4001a9fc <compute_BP1+0x144>
4001a9bc:	db 06 20 44 	ld  [ %i0 + 0x44 ], %f13
4001a9c0:	df 27 bf c8 	st  %f15, [ %fp + -56 ]
4001a9c4:	d1 27 bf d4 	st  %f8, [ %fp + -44 ]
4001a9c8:	d3 27 bf d0 	st  %f9, [ %fp + -48 ]
4001a9cc:	d5 27 bf d8 	st  %f10, [ %fp + -40 ]
4001a9d0:	d9 27 bf dc 	st  %f12, [ %fp + -36 ]
4001a9d4:	db 27 bf e0 	st  %f13, [ %fp + -32 ]
4001a9d8:	40 00 02 49 	call  4001b2fc <sqrtf>

What I understand is that fsqrts might fail and in this case we would repeat the computation with newlib’s soft implementation. Am I right? Under which circumstances could it fail? Can we avoid it?

Best regards,
Alexis.

[mhjorth]

Hi! Are you using our binary rcc distribution or built the toolchain yourself? Any build flags (-mtune etc)? Is this generated from a single sqrtf() function call in C?

Just from looking at the disassembly, it looks like this code checks for a NaN result from fsqrts (as NaN is not equal to itself by definition) which should happen if the input to the square root is less than 0.

[jeandet]

Hi, I use a self built rcc since we had to modify spw init for dual link mode:
teamcity-docker-SolarOrbiter-LFR-agent/Dockerfile at master · jeandet/teamcity-docker-SolarOrbiter-LFR-agent · GitHub
regarding CPU related build flags, nothing except -mfix-b2bst LFR_Flight_Software/meson.build at R3.3 · LaboratoryOfPlasmaPhysics/LFR_Flight_Software · GitHub

NaN results was my guess, but I wondered if it could happen with allowed values(>=0) because I would expect libc’s sqrt to return the same result (NaN) with anything < 0.

[jeandet]

Maybe it is called to get ERRNO set?

[jeandet]

Yes it was errno

[mhjorth]

Good find, thanks for reporting back!