Electronic – AVR: why each Interrupt Vector occupies two instruction words

The answer is in avr/interrupt.h line 228 which decides to use jmp on ATMEGA and rjmp on other micros. Presumably you could override it back to rjmp as long as you make sure your ISR's address is within the allowable range for rjmp.

The documentation for avr/interrupt.h as mentioned in another answer explains how to mark the ISR as naked to avoid any function setup/teardown overhead. You could even write the function using inline assembly, but you can probably get exactly the instructions you want from C, too.

There are a couple of issues:

• Not all AVR commands take 1 clock to be executed: if you look at the back of the datasheet, it has the number of clocks it takes for each instruction to be executed. So, for example AND is a one-clock instruction, MUL (multiply) takes two clocks, while LPM (load program memory) is three, and CALL is 4. So, with respect to the instruction execution, it really depends on the instruction.
• 5 clocks to jump in and 5 clocks to return can be misleading. If you look at your disassembled code, you will find that in addition to the jump and RETI instructions, the compiler adds all sorts of other code, which also takes time. For instance you might need local variables which are created on the stack and must be popped off, etc. The best thing to do to see what's actually going on is to look at the disassembly.
• Lastly, remember that while you are in your ISR routine, your interrupts are not triggering. This means that you will not be able to get the kind of performance you are looking for from your logic analyzer, unless you know that your signal levels change at intervals longer than it takes to service your interrupt. To be clear, once you calculate the time it takes for your ISR to execute, this gives you an upper limit of how quickly you can capture one signal. If you need to capture two signals, then you start runnning into trouble. To be overly detailed about this consider the following scenario:

If x is the time it takes to service your interrupt, then signal B will never be captured.

If we take your ISR code, stick it into an ISR routine (I used ISR(PCINT0_vect)) routine, declare all the variables volatile, and compile for ATmega168P, the disassembled code looks as follows (see @jipple's answer for more info) before we get to the code that "does something"; in orther words the prologue to your ISR is as follows:

  37                    .loc 1 71 0
  38                    .cfi_startproc
  39 0000 1F92              push r1
  40                .LCFI0:
  41                    .cfi_def_cfa_offset 3
  42                    .cfi_offset 1, -2
  43 0002 0F92              push r0
  44                .LCFI1:
  45                    .cfi_def_cfa_offset 4
  46                    .cfi_offset 0, -3
  47 0004 0FB6              in r0,__SREG__
  48 0006 0F92              push r0
  49 0008 1124              clr __zero_reg__
  50 000a 8F93              push r24
  51                .LCFI2:
  52                    .cfi_def_cfa_offset 5
  53                    .cfi_offset 24, -4
  54 000c 9F93              push r25
  55                .LCFI3:
  56                    .cfi_def_cfa_offset 6
  57                    .cfi_offset 25, -5
  58                /* prologue: Signal */
  59                /* frame size = 0 */
  60                /* stack size = 5 */
  61                .L__stack_usage = 5

so, PUSH x 5, in x 1, clr x 1. Not as bad as jipple's 32-bit vars, but still not nothing.

Some of this is necesary (expand the discussion in the comments). Obviosely, since the ISR routine can occur at any time, it must preseve the registers it uses, unless you know that no code where an interrupt can occur uses the same register as your interrupt routine. For example the following line in the disassembled ISR:

push r24

Is there because everything goes through r24: your pinc is loaded there before it goes into memory, etc. So you must have that first. __SREG__ is loaded into r0 and then pushed: if this could go through r24 then you could save yourself a PUSH

Some possible solutions:

• Use a tight polling loop as suggested by Kaz in the comments. This is probably going to be the fastest solution, whether you write the loop in C or assembly.
• Write your ISR in assembly: this way you can optimize the register usage in such a way that the fewest number of them need to be saved during the ISR.
• Declare your ISR routines ISR_NAKED, though this turns out to be more of a red herring solution. When you declare ISR routines ISR_NAKED, gcc does not generate prologue/epilogue code, and you are responsible for saving any registers your code modifies, as well as calling reti (return from an interrupt). Unfortunately, there is no way of using registers in avr-gcc C directly (obviously you can in assembly), however, what you can do is bind variables to specific registers with the register + asm keywords, like this: register uint8_t counter asm("r3");. If you do that, for the ISR you'll know what registers you are using in the ISR. The problem then is that there is no way to generate push and pop to save the used registers without inline assembly (cf. point 1). To ensure having to save fewer registers, you can also bind all the non-ISR variables to specific registers as well, however, no you run into a problem that gcc uses registers for shuffling data to and from memory. This means that unless you look at the disassembly you will not know what registers your main code uses. So if you are considering ISR_NAKED, you might as well write the ISR in assembly.