more.rst

More Performance

So far in all our examples we've been able to meet our timing goals by writing our code in the C programming language. The C compiler does a surprisingly good job at generating code, most the time. However there are times when very precise timing is needed and the compiler isn't doing it.

At these times you need to write in assembly language. This chapter introduces the PRU assembler and shows how to call assembly code from C. Detailing on how to program in assembly are beyond the scope of this text.

The following are resources used in this chapter.

Note

Resources

• PRU Optimizing C/C++ Compiler, v2.2, User's Guide
• PRU Assembly Language Tools User's Guide
• PRU Assembly Instruction User Guide

Calling Assembly from C

Problem

You have some C code and you want to call an assembly language routine from it.

Solution

You need to do two things, write the assembler file and modify the Makefile to include it. For example, let's write our own my_delay_cycles routine in in assembly. The intrinsic pass:[__]delay_cycles must be passed a compile time constant. Our new delay_cycles can take a runtime delay value.

:ref:`more_delay-test` is much like our other c code, but on line 10 we declare my_delay_cycles and then on lines 24 and 26 we'll call it with an argument of 1.

:download:`delay-test.pru0.c <code/delay-test.pru0.c>`

:ref:`more_delay` is the assembly code.

:download:`delay.pru0.asm <code/delay.pru0.asm>`

The Makefile has one addition that needs to be made to compile both :ref:`more_delay-test` and :ref:`more_delay`. If you look in the local Makefile you'll see:

:download:`Makefile <code/Makefile>`

This Makefle includes a common Makfile at /var/lib/cloud9/common/Makefile, this the Makefile you need to edit. Edit /var/lib/cloud9/common/Makefile and go to line 195.

$(GEN_DIR)/%.out: $(GEN_DIR)/%.o *$(GEN_DIR)/$(TARGETasm).o*
  @mkdir -p $(GEN_DIR)
  @echo 'LD   $^'
  $(eval $(call target-to-proc,$@))
  $(eval $(call proc-to-build-vars,$@))
  @$(LD) $@ $^ $(LDFLAGS)

Add *(GEN_DIR)/$(TARGETasm).o* as shown in bold above. You will want to remove this addition once you are done with this example since it will break the other examples.

The following will compile and run everything.

bone$ *config-pin P9_31 pruout*
bone$ *make TARGET=delay-test.pru0 TARGETasm=delay.pru0*
/var/lib/cloud9/common/Makefile:29: MODEL=TI_AM335x_BeagleBone_Black,TARGET=delay-test.pru0
-    Stopping PRU 0
-     copying firmware file /tmp/cloud9-examples/delay-test.pru0.out to /lib/firmware/am335x-pru0-fw
write_init_pins.sh
-    Starting PRU 0
MODEL   = TI_AM335x_BeagleBone_Black
PROC    = pru
PRUN    = 0
PRU_DIR = /sys/class/remoteproc/remoteproc1

The resulting output is shown in :ref:`more_my_delay_cycles`.

Output of my_delay_cycles()

Notice the on time is about 35ns and the off time is 30ns.

Discission

There is much to explain here. Let's start with :ref:`more_delay`.

Line-by-line of delay.pru0.asm

Line	Explanation
3	Declare my_delay_cycles to be global so the linker can find it.
4	Label the starting point for my_delay_cycles.
5	Label for our delay loop.
6	The first argument is passed in register r14. Page 111 of PRU Optimizing C/C++ Compiler, v2.2, User's Guide gives the argument passing convention. Registers r14 to r29 are used to pass arguments, if there are more arguments, the argument stack (r4) is used. The other register conventions are found on page 108. Here we subtract 1 from r14 and save it back into r14.
7	qbne is a quick branch if not equal.
9	Once we've delayed enough we drop through the quick branch and hit the jump. The upper bits of register r3 has the return address, therefore we return to the c code.

:ref:`more_my_delay_cycles` shows the on time is 35ns and the off time is 30ns. With 5ns/cycle this gives 7 cycles on and 6 off. These times make sense because each instruction takes a cycle and you have, set R30, jump to my_delay_cycles, sub, qbne, jmp. Plus the instruction (not seen) that initializes r14 to the passed value. That's a total of six instructions. The extra instruction is the branch at the bottom of the while loop.

Returning a Value from Assembly

Problem

Your assembly code needs to return a value.

Solution

R14 is how the return value is passed back. :ref:`more_test2` shows the c code.

:download:`delay-test2.pru0.c <code/delay-test2.pru0.c>`

:ref:`more_delay2` is the assembly code.

:download:`delay2.pru0.asm <code/delay2.pru0.asm>`

An additional feature is shown in line 4 of :ref:`more_delay2`. The .cdecls "delay-test2.pru0.c" says to include any defines from delay-test2.pru0.c In this example, line 6 of :ref:`more_test2` #defines TEST and line 12 of :ref:`more_delay2` reference it.

Using the Built-In Counter for Timing

Problem

I want to count how many cycles my routine takes.

Solution

Each PRU has a CYCLE register which counts the number of cycles since the PRU was enabled. They also have a STALL register that counts how many times the PRU stalled fetching an instruction. :ref:`more_cycle` shows they are used.

:download:`cycle.pru0.c <code/cycle.pru0.c>`

Discission

The code is mostly the same as other examples. cycle and stall end up in registers which we can read using prudebug. :ref:`more_cycle_lines` is the Line-by-line.

Line-by-line for cycle.pru0.c

Line	Explanation
4	Include needed to reference CYCLE and STALL.
16	Declaring cycle and stall. The compiler will optimize these and just keep them in registers. We'll have to look at the cycle.pru0.lst file to see where they are stored.
21	Enables CYCLE.
26	Reset CYCLE. It ignores the value assigned to it and always sets it to 0. cycle is on the right hand side to make the compiler give it its own register.
28, 29	Reads the CYCLE and STALL values into registers.

You can see where cycle and stall are stored by looking into :ref:`more_cycle_list0`.

:download:`cycle.pru0.lst <code/cycle.pru0.lst>`

Here the LDI32 instruction loads the address 0x22000 into r0. This is the offset to the CTRL registers. Later in the file we see :ref:`more_cycle_list1`.

:download:`cycle.pru0.lst <code/cycle.pru0.lst>`

The first LBBO takes the contents of r0 and adds the offset 12 to it and copies 4 bytes into r1. This points to CYCLE, so r1 has the contents of CYCLE.

The second LBBO does the same, but with offset 16, which points to STALL, thus STALL is now in r0.

Now fire up prudebug and look at those registers.

bone$ *sudo prudebug*
PRU0> *r*
r
r
Register info for PRU0
    Control register: 0x00000009
      Reset PC:0x0000  STOPPED, FREE_RUN, COUNTER_ENABLED, NOT_SLEEPING, PROC_DISABLED

    Program counter: 0x0012
      Current instruction: HALT

    R00: *0x00000005*    R08: 0x00000200    R16: 0x000003c6    R24: 0x00110210
    R01: *0x00000003*    R09: 0x00000000    R17: 0x00000000    R25: 0x00000000
    R02: 0x000000fc    R10: 0xfff4ea57    R18: 0x000003e6    R26: 0x6e616843
    R03: 0x0004272c    R11: 0x5fac6373    R19: 0x30203020    R27: 0x206c656e
    R04: 0xffffffff    R12: 0x59bfeafc    R20: 0x0000000a    R28: 0x00003033
    R05: 0x00000007    R13: 0xa4c19eaf    R21: 0x00757270    R29: 0x02100000
    R06: 0xefd30a00    R14: 0x00000005    R22: 0x0000001e    R30: 0xa03f9990
    R07: 0x00020024    R15: 0x00000003    R23: 0x00000000    R31: 0x00000000

So cycle is 3 and stall is 5. It must be one cycle to clear the GPIO and 2 cycles to read the CYCLE register and save it in the register. It's interesting there are 5 stall cycles.

If you switch the order of lines 30 and 31 you'll see cycle is 7 and stall is 2. cycle now includes the time needed to read stall and stall no longer includes the time to read cycle.

Xout and Xin - Transferring Between PRUs

Problem

I need to transfer data between PRUs quickly.

Solution

The pass:[__]xout() and pass:[__]xin() intrinsics are able to transfer up to 30 registers between PRU 0 and PRU 1 quickly. :ref:`more_xout` shows how xout() running on PRU 0 transfers six registers to PRU 1.

:download:`xout.pru0.c <code/xout.pru0.c>`

PRU 1 waits at line 41 until PRU 0 signals it. :ref:`more_xin` sends an interrupt to PRU 0 and waits for it to send the data.

:download:`xin.pru1.c <code/xin.pru1.c>`

Use prudebug to see registers R5-R10 are transferred from PRU 0 to PRU 1.

PRU0> *r*
Register info for PRU0
    Control register: 0x00000001
      Reset PC:0x0000  STOPPED, FREE_RUN, COUNTER_DISABLED, NOT_SLEEPING, PROC_DISABLED

    Program counter: 0x0026
      Current instruction: HALT

    R00: 0x00000012    *R08: 0xbbbbbbbb*    R16: 0x000003c6    R24: 0x00110210
    R01: 0x00020000    *R09: 0x87654321*    R17: 0x00000000    R25: 0x00000000
    R02: 0x000000e4    *R10: 0xcccccccc*    R18: 0x000003e6    R26: 0x6e616843
    R03: 0x0004272c    R11: 0x5fac6373    R19: 0x30203020    R27: 0x206c656e
    R04: 0xffffffff    R12: 0x59bfeafc    R20: 0x0000000a    R28: 0x00003033
    *R05: 0xdeadbeef*    R13: 0xa4c19eaf    R21: 0x00757270    R29: 0x02100000
    *R06: 0xaaaaaaaa*    R14: 0x00000005    R22: 0x0000001e    R30: 0xa03f9990
    *R07: 0x12345678*    R15: 0x00000003    R23: 0x00000000    R31: 0x00000000

PRU0> *pru 1*
pru 1
Active PRU is PRU1.

PRU1> *r*
r
Register info for PRU1
    Control register: 0x00000001
      Reset PC:0x0000  STOPPED, FREE_RUN, COUNTER_DISABLED, NOT_SLEEPING, PROC_DISABLED

    Program counter: 0x000b
      Current instruction: HALT

    R00: 0x00000100    *R08: 0xbbbbbbbb*    R16: 0xe9da228b    R24: 0x28113189
    R01: 0xe48cdb1f    *R09: 0x87654321*    R17: 0x66621777    R25: 0xddd29ab1
    R02: 0x000000e4    *R10: 0xcccccccc*    R18: 0x661f83ea    R26: 0xcf1cd4a5
    R03: 0x0004db97    R11: 0xdec387d5    R19: 0xa85adb78    R27: 0x70af2d02
    R04: 0xa90e496f    R12: 0xbeac3878    R20: 0x048fff22    R28: 0x7465f5f0
    *R05: 0xdeadbeef*    R13: 0x5777b488    R21: 0xa32977c7    R29: 0xae96b530
    *R06: 0xaaaaaaaa*    R14: 0xffa60550    R22: 0x99fb123e    R30: 0x52c42a0d
    *R07: 0x12345678*    R15: 0xdeb2142d    R23: 0xa353129d    R31: 0x00000000

Discussion

:ref:`more_zout_lines` shows the line-by-line for xout.pru0.c

xout.pru0.c Line-by-line

Line	Explanation
4	A different resource so PRU 0 can receive a signal from PRU 1.
9-16	dmemBuf holds the data to be sent to PRU 1. Each will be transferred to its corresponding register by xout().
21-22	Define the interrupts we're using.
27-28	Clear the interrupts.
31-36	Initialize dmemBuf with easy to recognize values.
40	Wait for PRU 1 to signal.
45	pass:[__]xout() does a direct transfer to PRU 1. Page 92 of PRU Optimizing C/C++ Compiler, v2.2, User's Guide shows how to use xout(). The first argument, 14, says to do a direct transfer to PRU 1. If the first argument is 10, 11 or 12, the data is transferred to one of three scratchpad memories that PRU 1 can access later. The second argument, 5, says to start transfering with register r5 and use as many registers as needed to transfer all of dmemBuf. The third argument, 0, says to not use remapping. (See the User's Guide for details.) The final argument is the data to be transferred.
48	Clear the interrupt so it can go again.

:ref:`more_xin_lines` shows the line-by-line for xin.pru1.c.

xin.pru1.c Line-by-line

Line	Explanation
8-15	Place to put the received data.
26	Signal PRU 0
30	Receive the data. The arguments are the same as xout(), 14 says to get the data directly from PRU 0. 5 says to start with register r5. dmemBuf is where to put the data.

If you really need speed, considering using pass:[__]xout() and pass:[__]xin() in assembly.

Copyright

:download:`copyright.c <code/copyright.c>`