You did not specify the kind of Arduino you are using but, based on another recent question of yours, I will assume it is an AVR-based Arduino. You cannot have an interrupt automatically triggered by a variable reaching zero. What you can do, instead, is trigger an interrupt by software, by toggling a pin that is configured both as an output and as an interrupt source: ```c++ const uint8_t int_pin = 2; // pin used for interrupt void setup() { digitalWrite(int_pin, LOW); pinMode(int_pin, OUTPUT); attachInterrupt(digitalPinToInterrupt(int_pin, reset, RISING)); } static inline void trigger_interrupt() { digitalWrite(int_pin, HIGH); digitalWrite(int_pin, LOW); } // Within loop(): if (!amp) trigger_interrupt(); ``` However, if you do that, you will discover that interrupts are slow. **Awfully slow** compared with a simple `if` test: 1. `digitalWrite()` itself is awfully slow. This can be overcome by using direct port access, but that is still two cycles for toggling the pin in each direction. 2. You will loose an extra cycle in the pin synchronizer. 3. Once the IRQ is raised, the CPU needs four cycles to prepare itself for servicing it. 4. The interrupt vector is a `jmp` instruction, which takes 3 cycles. 5. The ISR then has to save every single register it is going to use, including the status register. That takes two cycles per register. And there are quite a lot of registers to save... 6. This ISR, which is provided by the Arduino core, will then look for the interrupt handler you provided with `attachInterrupt()`. This also involves checking that the pointer to the handler is not zero (the kind of test you seem to be worried about). It then has to `call` (4 cycles) your handler, which will have to `ret`urn (4 cycles) to the ISR. You could avoid this indirection by defining the ISR yourself instead of relying on `attachInterrupt()`. 7. Once the job is done, all the saved registers have to be restored (2 cycles times many register), and the `reti` instruction (4 cycles) is issued in order to restore control to the interrupted program. 8. Last but not least, since `amp` will be modified in interrupt context, you will have to qualify it as `volatile`. This keyword prevents the compiler to perform optimizations that are unsafe in this situation, and you will loose _a lot_ of time to these missed optimization opportunities. In contrast, this line: ```c++ if (--amp == 0) amp = 10; ``` would be translated by the compiler into something like this: ```lisp dec amp ; --amp, as amp is likely already in a register brne 1f ; if (amp != 0) skip the following ldi amp, 10 ; amp = 10 1: ``` The whole sequence takes 3 cycles, irrespective of whether the `if` condition was true or false. That is at least one order of magnitude faster than the interrupt-based solution.