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Cortex A8 Instruction Cycle Timing

Note: This was originally posted on 17th March 2011 at http://forums.arm.com

Hi) sorry for bad English

I need to count latency for two instruction, and all I have is the arm cortex A 8 documantation(charter 16) !
but I have no idea how can do this work using that documantation(

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0 Etienne SOBOLE over 12 years ago

Note: This was originally posted on 11th April 2011 at http://forums.arm.com

The problem is due to your branch.

You can't simply expect that, if the branch is in the cache, it will take 1 cycle...
branch is more complex as it seem's.

Take this testing procédure. It will be more easy to understand the time taken by your program.
movw r0, #0x0500 @ you repeat your loop 83232000 times movt r0, #0x04F6 .loop: nop @ here is you nop nop nop nop nop nop nop nop smuad r1, r1, r1 @ you can be sure the ending code take 5 cycles. nop nop smuad r2, r2, r2 nop subs r0, r0, #1 smuad r3, r3, r3 bgt .loop bx lr

If you don't put any nop (I speak about your nop ! don't remove the nop after this smuad)
the program should take 0.52 s (this is logic because you beagleboard works at 800mhz and 5 * 83232000 ~= 400M cycles
every time you add 2 nop you program will take ~= 0.10 s more

you could have more readable result is you repeat you loop 80.000.000 times
in this case use

movw r0, #0xB400 movt r0, #0x04C4

instead of

movw r0, #0x0500 movt r0, #0x04F6
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0 Etienne SOBOLE over 12 years ago

Note: This was originally posted on 11th April 2011 at http://forums.arm.com

The problem is due to your branch.

You can't simply expect that, if the branch is in the cache, it will take 1 cycle...
branch is more complex as it seem's.

Take this testing procédure. It will be more easy to understand the time taken by your program.
movw r0, #0x0500 @ you repeat your loop 83232000 times movt r0, #0x04F6 .loop: nop @ here is you nop nop nop nop nop nop nop nop smuad r1, r1, r1 @ you can be sure the ending code take 5 cycles. nop nop smuad r2, r2, r2 nop subs r0, r0, #1 smuad r3, r3, r3 bgt .loop bx lr

If you don't put any nop (I speak about your nop ! don't remove the nop after this smuad)
the program should take 0.52 s (this is logic because you beagleboard works at 800mhz and 5 * 83232000 ~= 400M cycles
every time you add 2 nop you program will take ~= 0.10 s more

you could have more readable result is you repeat you loop 80.000.000 times
in this case use

movw r0, #0xB400 movt r0, #0x04C4

instead of

movw r0, #0x0500 movt r0, #0x04F6
Cancel
Vote up 0 Vote down

Cancel

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