LPC MicrocontrollersのトピックRe: Cycles needed to change an IO pin

Cycles needed to change an IO pin

lpcware — Wed, 15 Jun 2016 19:27:21 GMT

Content originally posted in LPCWare by vladimirro on Fri Nov 06 01:26:47 MST 2015
Hello all

I always wondered how many processor cycles does it take to change an IO pin in arm microcontrollers. For example I want to change an output from low to high. Will it be done in one cycle?

Re: Cycles needed to change an IO pin

lpcware — Wed, 15 Jun 2016 19:27:22 GMT

Content originally posted in LPCWare by R2D2 on Fri Nov 06 05:36:32 MST 2015

Quote: vladimirro
For example I want to change an output from low to high. Will it be done in one cycle?

No, you'll need >= 2 cycles, usually 3-4 :O

Re: Cycles needed to change an IO pin

lpcware — Wed, 15 Jun 2016 19:27:23 GMT

Content originally posted in LPCWare by wmues on Fri Nov 06 07:18:19 MST 2015
You may use bit banding in cortex-m controllers.

Re: Cycles needed to change an IO pin

lpcware — Wed, 15 Jun 2016 19:27:24 GMT

Content originally posted in LPCWare by R2D2 on Fri Nov 06 08:03:15 MST 2015

Quote: wmues
You may use bit banding in cortex-m controllers.

:D

We are talking about M0 here, aren't we?

Re: Cycles needed to change an IO pin

lpcware — Wed, 15 Jun 2016 19:27:24 GMT

Content originally posted in LPCWare by vladimirro on Fri Nov 06 10:26:04 MST 2015

Quote: R2D2

No, you'll need >= 2 cycles, usually 3-4 :O

What does it depend on?

Re: Cycles needed to change an IO pin

lpcware — Wed, 15 Jun 2016 19:27:25 GMT

Content originally posted in LPCWare by R2D2 on Sat Nov 07 16:28:29 MST 2015
It depends on your code and compiler...

Use the disassembly to see what's happening:

Sample: Set / clear GPIO3_0

Quote:
[color=#c00]0x00000328: 0x4b03 main+40 ldr r3, [pc, #12]       ; (0x338 <main+56>)
0x0000032a: 0x2101 main+42 movs r1, #1
0x0000032c: 0x6059 main+44 str r1, [r3, #4][/color]
[color=#03f]25                                            LPC_GPIO[3].DATA[1 << 0] = 0 << 0;
0x0000032e: 0x2200 main+46 movs r2, #0
0x00000330: 0x605a main+48 str r2, [r3, #4][/color]
[color=#090]26                         LPC_GPIO[3].DATA[1 << 0] = 1 << 0;
0x00000332: 0x6059 main+50 str r1, [r3, #4]
27                         LPC_GPIO[3].DATA[1 << 0] = 0 << 0;
0x00000334: 0x605a main+52 str r2, [r3, #4][/color]
28                         }
0x00000336: 0xe7f7 main+54 b.n 0x328 <main+40>
0x00000338: 0x0000 main+56 movs r0, r0
0x0000033a: 0x5003 main+58 str r3, [r0, r0]

In main+50 and main+52 you can see the fastest option: one store instruction. That's requiring 2 cycles :)

In main+46 - main+48 there's an additional move instruction to load 0 in r2. So that's requiring 2 (store) + 1 (move) = 3 cycles.

And in main+40 - main+44 an additional load instruction and adding even more cycles.

In general the compiler is trying to reduce instructions and reuse register values. If that's not possible, registers have to be loaded again and more cycles are needed :(