51 External interruptions to learning-EEWORLD

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When learning about external interrupts, because I didn’t understand them, I just pasted the program to execute them.

void ISR0_Key() interrupt 0 using 1

{

Speak1=~Speak1;

}

void ISR1_Key() interrupt 2 using 2

{

Speak2=~Speak2;

}

void Main()

{

EA=1; //Global disconnect

EX0=1; //External interrupt 0 is on

EX1=1; //External interrupt 1 is on

IT0=1; //Edge trigger, 0 means low level trigger

IT1=0; //Edge trigger, 0 means low level trigger

while(1);

}

The execution results seem a bit confusing. I don't quite understand the difference between edge triggering and level triggering. I read about it online and I can understand it, but the actual execution is different.

1. When the keyboard is pressed in the first scan, the result is the same as what we said before. Whether it is a low level trigger or a falling edge trigger, the event will be triggered. Here, the indicator light is on when it is triggered. Then the program continues to run...

2. When the program continues to run and scans the keyboard for the second time, if the key has not been released, the result will be different.

Low level trigger: The trigger condition is still at a low level. If the trigger condition is met, the event is executed and the indicator light status changes. At this time, the indicator light changes from on to off.

Falling edge trigger: The trigger condition is still at a low level. If the falling edge trigger condition is not met, the event will not be executed and the indicator light will remain on.

Haha, let’s study it tomorrow!

There are also some problems with the interrupt handler.

The interrupt service routine uses the interrupt keyword and the interrupt number (0 to 31) to determine which interrupt the interrupt service routine is handling.
The using keyword is used to specify the register group used by the interrupt service routine. The usage is: using followed by a number from 0 to 3, corresponding to 4 groups of working registers. Once the working register group is specified, the default working register group will not be pushed onto the stack, which will save 32 processing cycles, because both stacking and popping require 2 processing cycles.

Today, I only tried the interrupt entry method. Tomorrow, I will try the query method, which should be the same as the timer.

Reference address：51 External interruptions to learning

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