4 years ago.

NUCLEO-F746ZG InterruptIn problem

During my attempts to debug the following test code, (The interrupt for the button will not fire after adding the next interrupts whether set or not)

#include "mbed.h"

DigitalOut led1(LED1);
DigitalOut led2(LED2);
DigitalOut led3(LED3);
InterruptIn quadButton(PB_9);
InterruptIn quadA(PC_9);
InterruptIn quadB(PC_8);

void int1() {
    led1 =! led1;
}

void int2() {
    led2 =! led2;    
}

void int3() {
    led3 =! led3;    
}

int main() {
            
    quadButton.rise(&int1);
    quadButton.mode(PullUp);
    
    quadA.rise(&int2);
    quadA.mode(PullUp);
    
    quadB.rise(&int3);
    quadB.mode(PullUp);
       
    while(true) {
    }
}

I've been reading the initialisation code that is executed as part of the InteruptIn constructor, and during TARGET_STM32F7/gpio_irq_api.c the irqs are assigned ect, but what I can't understand is how 2 pins on different ports with the same number can be used together.

using pins PB_9 and PC_9, in gpio_irq_init() the switch statement is executed using 9 (returned from STM_PIN(pin)) so both pins are assigned to irq5, then the irq_index (5) is used to select a channel, so the pins are assigned the same channel, the channel index is selected using a lookup table returning 4 for both, then the data is stored into that index, for both, overwriting the previous data. This doesn't make much sense so I assume I've missed something.

Anyway this is a very long winded way of asking any idea whats going on with the interrupts?

it appears to be a hardware limitation, each interrupt line can only support 1 port at a time

posted by Chris Pepper 05 Aug 2016
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