I wrote an an idle hook shown here
void vApplicationIdleHook( void )
{
asm("nop");
P1OUT &= ~0x01;//go to sleep lights off!
LPM3;// LPM Mode - remove to make debug a little easier...
asm("nop");
}
That should cause the LED to turn off, and MSP430 to go to sleep when there is nothing to do. I turn the LED on during some tasks.
I also made sure to modify the sleep mode bit in the SR upon exit of any interrupt that could possibly wake the MCU (with the exception of the scheduler tick isr in portext.s43. The macro in iar is
__bic_SR_register_on_exit(LPM3_bits); // Exit Interrupt as active CPU
However, it seems as though putting the MCU to sleep causes some irregular behavior. The led stays on always, although when i scope it, it will turn off for a couple instructions cycles when ever i wake the mcu via one of the interrupts (UART), and then turn back on.
If I comment out the LPM3 instruction, things go as planned. The led stays off for most of the time and only comes on when a task is running.
I am using a MSP4f305438
Any ideas?
Perhaps the problem is the call __bic_SR_register_on_exit(LPM3_bits). This macro changes the LPM bits in the stacked SR, so it must know where to find the saved SR on the stack. I believe that __bic_SR_register_on_exit() is designed for the standard interrupt stack frame generated by the compiler when you use the __interrupt directive. However, a preemptive RTOS, like FreeRTOS, uses its own stack frame typically bigger than the stack frame generated by the compiler, because an RTOS must store the complete context. In this case __bic_SR_register_on_exit() called from an ISR might not find the SR on the stack. Worse, it probably corrupts some other saved register value on the stack.
For a preemptive kernel I would not call __bic_SR_register_on_exit() from the ISRs. The consequence is that the idle callback is called only once and never again, because every time the RTOS performs a context switch back to the idle task the side effect is restoring the SR with the LPM bits turned on. This causes a sleep mode (which is what you want), but your LED won't get toggled.
Miro Samek
state-machine.com
Related
I'm working with TIM2 on the STM32L068K which is a Cortex M0 processor. When I write to the timer enable bit, all the interrupt flags immediately get set. This in itself is a known issue and apparently endemic to the processor design based on the online commentary I've read.
I can clear out the interrupt flags by writing to the status register, but the problem is that the NVIC pending IRQ bit for this source (#15) is also set. This means that the second I execute cpsie i I get vectored to the ISR for source #15 (confirmed by seeing that this is the reported source in IPSR). I've tried multiple techniques for writing to NVIC_ICPR, but the bit remains set. As one example of many things I've tried, check out this site : https://www.sciencedirect.com/topics/engineering/pending-interrupt. I've also tried the CMSIS calls to no good effect. Do writes to this register only work in handler mode, not thread mode? And if so, how then can you stop a spurious interrupt from happening? Is it possible to manually enable handler mode without triggering an exception?
Note that this website does say "If the interrupt source generates an interrupt request continuously (level output), then the pending status could remain high even if you try to clear it at the NVIC." I wouldn't expect the TIM2 IRQ to fall into this category as it should only be triggering when the count reaches zero, which is not happening here, and the interrupt flags for it have already been cleared anyway.
I'm using the blue pill, and trying to figure out interrupts. I have an interrupt handler:
void __attribute__ ((interrupt ("TIM4_IRQHandler"))) myhandler()
{
puts("hi");
TIM4->EGR |= TIM_EGR_UG; // send an update even to reset timer and apply settings
TIM4->SR &= ~0x01; // clear UIF
TIM4->DIER |= 0x01; // UIE
}
I set up the timer:
RCC_APB1ENR |= RCC_APB1ENR_TIM4EN;
TIM4->PSC=7999;
TIM4->ARR=1000;
TIM4->EGR |= TIM_EGR_UG; // send an update even to reset timer and apply settings
TIM4->EGR |= (TIM_EGR_TG | TIM_EGR_UG);
TIM4->DIER |= 0x01; // UIE enable interrupt
TIM4->CR1 |= TIM_CR1_CEN;
My timer doesn't seem to activate. I don't think I've actually enabled it though. Have I??
I see in lots of example code commands like:
NVIC_EnableIRQ(USART1_IRQn);
What is actually going in NVIC_EnableIRQ()?
I've googled around, but I can't find actual bare-metal code that's doing something similar to mine.
I seem to be missing a crucial step.
Update 2020-09-23 Thanks to the respondents to this question. The trick is to set the bit for the interrupt number in an NVIC_ISER register. As I pointed out below, this doesn't seem to be mentioned in the STM32F101xx reference manual, so I probably would never have been able to figure this out on my own; not that I have any real skill in reading datasheets.
Anyway, oh joy, I managed to get interrupts working! You can see the code here: https://github.com/blippy/rpi/tree/master/stm32/bare/04-timer-interrupt
Even if you go bare metal, you might still want to use the CMSIS header files that provide declarations and inline version of very basic ARM Cortex elements such NVIC_EnableIRQ.
You can find NVIC_EnableIRQ at https://github.com/ARM-software/CMSIS_5/blob/develop/CMSIS/Core/Include/core_cm3.h#L1508
It's defined as:
#define NVIC_EnableIRQ __NVIC_EnableIRQ
__STATIC_INLINE void __NVIC_EnableIRQ(IRQn_Type IRQn)
{
if ((int32_t)(IRQn) >= 0)
{
__COMPILER_BARRIER();
NVIC->ISER[(((uint32_t)IRQn) >> 5UL)] = (uint32_t)(1UL << (((uint32_t)IRQn) & 0x1FUL));
__COMPILER_BARRIER();
}
}
If you want to, you can ignore __COMPILER_BARRIER(). Previous versions didn't use it.
The definition is applicable to Cortex M-3 chips. It's different for other Cortex versions.
With the libraries is still considered bare metal. Without operating system, but anyway, good that you have a desire to learn at this level. Someone has to write the libraries for others.
I was going to do a full example here, (it really takes very little code to do this), but will take from my code for this board that uses timer1.
You obviously need the ARM documentation (technical reference manual for the cortex-m3 and the architectural reference manual for armv7-m) and the data sheet and reference manual for this st part (no need for programmers manual from either company).
You have provided next to no information related to making the part work. You should never dive right into a interrupt, they are advanced topics and you should poll your way as far as possible before finally enabling the interrupt into the core.
I prefer to get a uart working then use that to watch the timer registers when the roll over, count, etc. Then see/confirm the status register fired, learn/confirm how to clear it (sometimes it is just a clear on read).
Then enable it into the NVIC and by polling see the NVIC sees it, and that you can clear it.
You didn't show your vector table this is key to getting your interrupt handler working. Much less the core booting.
08000000 <_start>:
8000000: 20005000
8000004: 080000b9
8000008: 080000bf
800000c: 080000bf
...
80000a0: 080000bf
80000a4: 080000d1
80000a8: 080000bf
...
080000b8 <reset>:
80000b8: f000 f818 bl 80000ec <notmain>
80000bc: e7ff b.n 80000be <hang>
...
080000be <hang>:
80000be: e7fe b.n 80000be <hang>
...
080000d0 <tim1_handler>:
The first word loads the stack pointer, the rest are vectors, the address to the handler orred with one (I'll let you look that up).
In this case the st reference manual shows that interrupt 25 is TIM1_UP at address 0x000000A4. Which mirrors to 0x080000A4, and that is where the handler is in my binary, if yours is not then two things, one you can use VTOR to find an aligned space, sometimes sram or some other flash space that you build for this and point there, but your vector table handler must have the proper pointer or your interrupt handler won't run.
volatile unsigned int counter;
void tim1_handler ( void )
{
counter++;
PUT32(TIM1_SR,0);
}
volatile isn't necessarily the right way to share a variable between interrupt handler and foreground task, it happens to work for me with this compiler/code, you can do the research and even better, examine the compiler output (disassemble the binary) to confirm this isn't a problem.
ra=GET32(RCC_APB2ENR);
ra|=1<<11; //TIM1
PUT32(RCC_APB2ENR,ra);
...
counter=0;
PUT32(TIM1_CR1,0x00001);
PUT32(TIM1_DIER,0x00001);
PUT32(NVIC_ISER0,0x02000000);
for(rc=0;rc<10;)
{
if(counter>=1221)
{
counter=0;
toggle_led();
rc++;
}
}
PUT32(TIM1_CR1,0x00000);
PUT32(TIM1_DIER,0x00000);
A minimal init and runtime for tim1.
Notice that the NVIC_ISER0 is bit 25 that is set to enable interrupt 25 through.
Well before trying this code, I polled the timer status register to see how it works, compare with docs, clear the interrupt per the docs. Then with that knowledge confirmed with the NVIC_ICPR0,1,2 registers that it was interrupt 25. As well as there being no other gates between the peripheral and the NVIC as some chips from some vendors may have.
Then released it through to the core with NVIC_ISER0.
If you don't take these baby steps and perhaps you have already, it only makes the task much worse and take longer (yes, sometimes you get lucky).
TIM4 looks to be interrupt 30, offset/address 0x000000B8, in the vector table. NVIC_ISER0 (0xE000E100) covers the first 32 interrupts so 30 would be in that register. If you disassemble the code you are generating with the library then we can see what is going on, and or look it up in the library source code (as someone already did for you).
And then of course your timer 4 code needs to properly init the timer and cause the interrupt to fire, which I didn't check.
There are examples, you need to just keep looking.
The minimum is
vector in the table
set the bit in the interrupt set enable register
enable the interrupt to leave the peripheral
fire the interrupt
Not necessarily in that order.
I've read (and studied) about Interrupt Handling.
What I always fail to understand, is how do we know where to return to (PC / IP) from the Interrupt Handler.
As I understand it:
An Interrupt is caused by a device (say the keyboard)
The relevant handler is called - under the running process. That is, no context switch to the OS is performed.
The Interrupt Handler finishes, and passes control back to the running application.
The process depicted above, which is my understanding of Interrupt Handling, takes place within the current running process' context. So it's akin to a method call, rather than to a context switch.
However, being that we didn't actually make the CALL to the Interrupt Handler, we didn't have a chance to push the current IP to the stack.
So how do we know where to jump back from an Interrupt. I'm confused.
Would appreciate any explanation, including one-liners that simply point to a good pdf/ppt addressing this question specifically.
[I'm generally referring to above process under Linux and C code - but all good answers are welcomed]
It's pretty architecture dependent.
On Intel processors, the interrupt return address is pushed on the stack when an interrupt occurs. You would use an iret instruction to return from the interrupt context.
On ARM, an interrupt causes a processor mode change (to the INT, FIQ, or SVC mode, for example), saving the current CPSR (current program status register) into the SPSR (saved program status register), putting the current execution address into the new mode's LR (link register), and then jumping to the appropriate interrupt vector. Therefore, returning from an interrupt is done by moving the SPSR into the CPSR and then jumping to an address saved in LR - usually done in one step with a subs or movs instruction:
movs pc, lr
When an interrupt is triggered, the CPU pushes several registers onto the stack, including the instruction pointer (EIP) of the code that was executing before the interrupt. You can put iret and the end of your ISR to pop these values, and restore EIP (as well as CS, EFLAGS, SS and ESP).
By the way, interrupts aren't necessarily triggered by devices. In Linux and DOS, user space programs use interrupts (via int) to make system calls. Some kernel code uses interrupts, for example intentionally triple faulting in order to force a shutdown.
The interrupt triggering mechanism in the CPU pushes the return address on the stack (among other things).
I was reading up on interrupts. It is possible to suspend non-critical interrupts via a special interrupt mask. This is called interrupt masking. What i dont know is when/why you might want to or need to temporarily suspend interrupts? Possibly Semaphores, or programming in a multi-processor environment?
The OS does that when it prepares to run its own "let's orchestrate the world" code.
For example, at some point the OS thread scheduler has control. It prepares the processor registers and everything else that needs to be done before it lets a thread run so that the environment for that process and thread is set up. Then, before letting that thread run, it sets a timer interrupt to be raised after the time it intends to let the thread have on the CPU elapses.
After that time period (quantum) has elapsed, the interrupt is raised and the OS scheduler takes control again. It has to figure out what needs to be done next. To do that, it needs to save the state of the CPU registers so that it knows how to undo the side effects of the code it executes. If another interrupt is raised for any reason (e.g. some async I/O completes) while state is being saved, this would leave the OS in a situation where its world is not in a valid state (in effect, saving the state needs to be an atomic operation).
To avoid being caught in that situation, the OS kernel therefore disables interrupts while any such operations that need to be atomic are performed. After it has done whatever needs doing and the system is in a known state again, it reenables interrupts.
I used to program on an ARM board that had about 10 interrupts that could occur. Each particular program that I wrote was never interested in more than 4 of them. For instance there were 2 timers on the board, but my programs only used 1. I would mask the 2nd timer's interrupt. If I didn't mask that timer, it might have been enabled and continued making interrupts which would slow down my code.
Another example was that I would use the UART receive REGISTER full interrupt and so would never need the UART receive BUFFER full interrupt to occur.
I hope this gives you some insight as to why you might want to disable interrupts.
In addition to answers already given, there's an element of priority to it. There are some interrupts you need or want to be able to respond to as quickly as possible and others you'd like to know about but only when you're not so busy. The most obvious example might be refilling the write buffer on a DVD writer (where, if you don't do so in time, some hardware will simply write the DVD incorrectly) versus processing a new packet from the network. You'd disable the interrupt for the latter upon receiving the interrupt for the former, and keep it disabled for the duration of filling the buffer.
In practise, quite a lot of CPUs have interrupt priority built directly into the hardware. When an interrupt occurs, the disabled flags are set for lesser interrupts and, often, that interrupt at the same time as reading the interrupt vector and jumping to the relevant address. Dictating that receipt of an interrupt also implicitly masks that interrupt until the end of the interrupt handler has the nice side effect of loosening restrictions on interrupting hardware. E.g. you can simply say that signal high triggers the interrupt and leave the external hardware to decide how long it wants to hold the line high for without worrying about inadvertently triggering multiple interrupts.
In many antiquated systems (including the z80 and 6502) there tends to be only two levels of interrupt — maskable and non-maskable, which I think is where the language of enabling or disabling interrupts comes from. But even as far back as the original 68000 you've got eight levels of interrupt and a current priority level in the CPU that dictates which levels of incoming interrupt will actually be allowed to take effect.
Imagine your CPU is in "int3" handler now and at that time "int2" happens and the newly happened "int2" has a lower priority compared with "int3". How would we handle with this situation?
A way is when handling "int3", we are masking out other lower priority interrupters. That is we see the "int2" is signaling to CPU but the CPU would not be interrupted by it. After we finishing handling the "int3", we make a return from "int3" and unmasking the lower priority interrupters.
The place we returned to can be:
Another process(in a preemptive system)
The process that was interrupted by "int3"(in a non-preemptive system or preemptive system)
An int handler that is interrupted by "int3"， say int1's handler.
In cases 1 and 2, because we unmasked the lower priority interrupters and "int2" is still signaling the CPU: "hi, there is a something for you to handle immediately", then the CPU would be interrupted again, when it is executing instructions from a process, to handle "int2"
In case 3, if the priority of “int2” is higher than "int1", then the CPU would be interrupted again, when it is executing instructions from "int1"'s handler, to handle "int2".
Otherwise, "int1"'s handler is executed without interrupting (because we are also masking out the interrupters with priority lower then "int1" ) and the CPU would return to a process after handling the “int1” and unmask. At that time "int2" would be handled.
In my systems programming class we are working on a small, simple hobby OS. Personally I have been working on an ATA hard disk driver. I have discovered that a single line of code seems to cause a fault which then immediately reboots the system. The code in question is at the end of my interrupt service routine for the IDE interrupts. Since I was using the IDE channels, they are sent through the slave PIC (which is cascaded through the master). Originally my code was only sending the end-of-interrupt byte to the slave, but then my professor told me that I should be sending it to the master PIC as well.
SO here is my problem, when I un-comment the line which sends the EOI byte to the master PIC, the systems triple faults and then reboots. Likewise, if I leave it commented the system stays running.
_outb( PIC_MASTER_CMD_PORT, PIC_EOI ); // this causes (or at least sets off) a triple fault reboot
_outb( PIC_SLAVE_CMD_PORT, PIC_EOI );
Without seeing the rest of the system, is it possible for someone to explain what could possibly be happening here?
NOTE: Just as a shot in the dark, I replaced the _outb() call with another _outb() call which just made sure that the interrupts were enable for the IDE controller, however, the generated assembly would have been almost identical. This did not cause a fault.
*_outb() is a wrapper for the x86 OUTB instruction.
What is so special about my function to send EOI to the master PIC that is an issue?
I realize without seeing the code this may be impossible to answer, but thanks for looking!
Triple faults usually point to a stack overflow or odd stack pointer. When a fault or interrupt occurs, the system immediately tries to push some more junk onto the stack (before invoking the fault handler). If the stack is hosed, this will cause another fault, which then tries to push more stuff on the stack, which causes another fault. At this point, the system gives up on you and reboots.
I know this because I actually have a silly patent (while working at Dell about 20 years ago) on a way to cause a CPU reset without external hardware (used to be done through the keyboard controller):
MOV ESP,1
PUSH EAX ; triple fault and reset!
An OUTB instruction can't cause a fault on its own. My guess is you are re-enabling an interrupt, and the interrupt gets triggered while something is wrong with your stack.
When you re-enable the PIC, are you doing it with the CPU's interrupt flag set, or cleared (ie. are you doing it sometime after a CLI opcode, or, sometime after an STI opcode)?
Assuming that the CPU's interrupt flag is enabled, your act of re-enabling the PIC allows any pending interrupts to reach the CPU: which would interrupt your code, dispatch to a vector specified by the IDT, etc.
So I expect that it's not your opcode that's directly causing the fault: rather, what's faulting is code that's run as the result of an interrupt which happens as a result of your re-enabling the PIC.