Why doesn't gcc handle volatile register? - gcc

Question

I'm working on a timing loop for the AVR platform where I'm counting down a single byte inside an ISR. Since this task is a primary function of my program, I'd like to permanently reserve a processor register so that the ISR doesn't have to hit a memory barrier when its usual code path is decrement, compare to zero, and reti.
The avr-libc docs show how to bind a variable to a register, and I got that working without a problem. However, since this variable is shared between the main program (for starting the timer countdown) and the ISR (for actually counting and signaling completion), it should also be volatile to ensure that the compiler doesn't do anything too clever in optimizing it.
In this context (reserving a register across an entire monolithic build), the combination volatile register makes sense to me semantically, as "permanently store this variable in register rX, but don't optimize away checks because the register might be modified externally". GCC doesn't like this, however, and emits a warning that it might go ahead and optimize away the variable access anyway.
The bug history of this combination in GCC suggests that the compiler team is simply unwilling to consider the type of scenario I'm describing and thinks it's pointless to provide for it. Am I missing some fundamental reason why the volatile register approach is in itself a Bad Idea, or is this a case that makes semantic sense but that the compiler team just isn't interested in handling?

Answer 1

The semantics of volatile are not exactly as you describe "don't optimize away checks because the register might be modified externally" but are actually more narrow: Try to think of it as "don't cache the variable's value from RAM in a register".
Seen this way, it does not make any sense to declare a register as volatile because the register itself cannot be 'cached' and therefore cannot possibly be inconsistent with the variable's 'actual' value.
The fact that read accesses to volatile variables are usually not optimzed away is merely a side effect of the above semantics, but it's not guaranteed.
I think GCC should assume by default that a value in a register is 'like volatile' but I have not verified that it actually does so.
Edit:
I just did a small test and found:
avr-gcc 4.6.2 does not treat global register variables like volatiles with respect to read accesses, and
the Naggy extension for Atmel Studio detects an error in my code: "global register variables are not supported".
Assuming that global register variables are actually considered "unsupported" I am not surprised that gcc treats them just like local variables, with the known implications.
My test code looks like this:
uint8_t var;
volatile uint8_t volVar;
register uint8_t regVar asm("r13");
#define NOP asm volatile ("nop\r\n":::)
int main(void)
{
var = 1; // <-- kept
if ( var == 0 ) {
NOP; // <-- optimized away, var is not volatile
}
volVar = 1; // <-- kept
if ( volVar == 0 ) {
NOP; // <-- kept, volVar *is* volatile
}
regVar = 1; // <-- optimized away, regVar is treated like a local variable
if ( regVar == 0 ) {
NOP; // <-- optimized away consequently
}
for(;;){}
}

Answer 2

The reason you would use the volatile keyword on AVR variables is to, as you said, avoid the compiler optimizing access to the variable. The question now is, how does this happen though?
A variable has two places it can reside. Either in the general purpose register file or in some location in RAM. Consider the case where the variable resides in RAM. To access the latest value of the variable, the compiler loads the variable from RAM, using some form of the ld instruction, say lds r16, 0x000f. In this case, the variable was stored in RAM location 0x000f and the program made a copy of this variable in r16. Now, here is where things get interesting if interrupts are enabled. Say that after loading the variable, the following occurs inc r16, then an interrupt triggers and its corresponding ISR is run. Within the ISR, the variable is also used. There is a problem, however. The variable exists in two different versions, one in RAM and one in r16. Ideally, the compiler should use the version in r16, but this one is not guaranteed to exist, so it loads it from RAM instead, and now, the code does not operate as needed. Enter then the volatile keyword. The variable is still stored in RAM, however, the compiler must ensure that the variable is updated in RAM before anything else happens, thus the following assembly may be generated:
cli
lds r16, 0x000f
inc r16
sei
sts 0x000f, r16
First, interrupts are disabled. Then, the the variable is loaded into r16. The variable is increased, interrupts are enabled and then the variable is stored. It may appear confusing for the global interrupt flag to be enabled before the variable is stored back in RAM, but from the instruction set manual:
The instruction following SEI will be executed before any pending interrupts.
This means that the sts instruction will be executed before any interrupts trigger again, and that the interrupts are disabled for the minimum amount of time possible.
Consider now the case where the variable is bound to a register. Any operations done on the variable are done directly on the register. These operations, unlike operations done to a variable in RAM, can be considered atomic, as there is no read -> modify -> write cycle to speak of. If an interrupt triggers after the variable is updated, it will get the new value of the variable, since it will read the variable from the register it was bound to.
Also, since the variable is bound to a register, any test instructions will utilize the register itself and will not be optimized away on the grounds the compiler may have a "hunch" it is a static value, given that registers by their very nature are volatile.
Now, from experience, when using interrupts in AVR, I have sometimes noticed that the global volatile variables never hit RAM. The compiler kept them on the registers all the time, bypassing the read -> modify -> write cycle alltogether. This was due, however, to compiler optimizations, and it should not be relied on. Different compilers are free to generate different assembly for the same piece of code. You can generate a disassembly of your final file or any particular object files using the avr-objdump utility.
Cheers.

Answer 3

Reserving a register for one variable for a complete compilation unit is probably too restrictive for a compiler's code generator. That is, every C routine would have to NOT use that register.
How do you guarantee that other called routines do NOT use that register once your code goes out of scope? Even stuff like serial i/o routines would have to NOT use that reserved register. Compilers do NOT recompile their run-time libraries based on a data definition in a user program.
Is your application really so time sensitive that the extra delay for bringing memory up from L2 or L3 can be detected? If so, then your ISR might be running so frequently that the required memory location is always available (i.e. it doesn't get paged back down thru the cache) and thus does NOT hit a memory barrier (I assume by memory barrier you are referring to how memory in a cpu really operates, through caching, etc.). But for this to really be true the up would have to have a fairly large L1 cache and the ISR would have to run at a very high frequency.
Finally, sometimes an application's requirements make it necessary to code it in ASM in which case you can do exactly what you are requesting!