Please help me write a program to monitor the status of 2 buttons. Is it really that difficult? I just can't figure it out!

yfpc2006

Please help me write a program to monitor the status of 2 buttons. Is it really that difficult? I just can't figure it out! [Copy link]

 

A car window lifter: A single-pole double-throw switch, KEY1, KEY2; Two output points OUT1, OUT2 No press: KEY1=1; KEY2=1; Press lightly: KEY1=0; KEY2=1; Press hard: KEY1=1; KEY2=0; The switching process from light press to hard press, or from hard press to light press: KEY1=1; KEY2=1; takes at least several hundred milliseconds. Program OUT1, OUT2 output requirements: When KEY1=0, OUT1=1; When KEY2=0, OUT1=1; OUT2=1; In the process from light press to heavy press, keep: OUT1=1; cannot be interrupted In the process from heavy press to light press, keep: OUT1=1; cannot be interrupted Not pressed: OUT1=0; OUT2=0; I can't figure out how to write the logic~ Please give me some advice, masters!

蓝雨夜

Is it not enough? If you press lightly, will you press harder? Is it not enough to press lightly?

y909334873

It should be enough to determine the key operation by judging the duration of the key state, right?                                   

yfpc2006

Blue Rain Night posted on 2019-2-14 14:37 Is it not enough? If you press lightly, will you definitely press hard? Is a light press not enough?

There are three levels in total: no press, light press, and heavy press. Please help!

yfpc2006

y909334873 posted on 2019-2-14 14:41 It should be enough to determine the key operation by judging the duration of the key state?

Newbies don’t know how to do this!

蓝雨夜

yfpc2006 posted on 2019-2-14 20:05 There are three gears in total: no press, light press, and hard press. Please help!

There are three gears in total: no press, light press, and hard press. What if you only press lightly and then don't press again?

小苹果单片机

I am also a novice, but you can see if this is right. You can check it according to this logic: if(KEY1 = 1 && key2 = 1) //No action in the initial state{ delay 1 s; if(KEY1 =1 && KEY2 =1) //Delay for one second if the key is not pressed{ OUT1= 0; OUT2 = 0； } else break; } else if(KEY1 = 0 || KEY2 = 0) //A key is pressed{ if(KEY1 =0 && KEY2 = 1) //If it is a light press{ { OUT1= 1;OUT2 = 0;} delay 1s; //Delay for one secondif(KEY1 =1 && KEY2= 0) //Judge whether the state is switchedOUT2 =1; else break; } else if(KEY1 =1 && KEY2 = 0) //If it is pressed hard{ OUT1 =1; OUT2 =1; delay 1s; if(KEY1 =0 && KEY2 = 1) //Judge state switching { OUT1 = 1; OUT2 = 0; } else break; } else //If KEY1 = 0 KEY2 = 0, then invalid break; } I don't know if it's right, you can take a look. This is probably the logic of the meaning.

懒猫爱飞

Sort out the state and write a simple state machine

邱海涛qht

Do you mean short press and long press when you say light press and heavy press? Or can you please post the picture of the SPDT switch you mentioned for reference?

philipchiu

The description is not clear enough, please post the switch information or switch model

取名好难

Operation key value status Output status Not pressed: KEY1=1; KEY2=1; 0 OUT1=0; OUT2=0; Press lightly: KEY1=0; KEY2=1; 1 OUT1=1; OUT2=0; Switch between light and heavy: KEY1=1; KEY2=1; 2 OUT1=1; OUT2=0; Press hard: KEY1=1; KEY2=0; 3 OUT1=1; OUT2=1; According to the original poster's description, is the status divided in this way? If so, you can first judge the status of 0-3, and then operate the output according to the status; the judgment from light press to release requires a delay judgment of "several hundred milliseconds";

hujj

The original poster should divide this problem into two parts, one is to obtain the key status, and the other is to process according to the key status, so that the logic may be clearer. If the original poster's heavy press and light press mean long press and short press, then you can assign key1=0 first, and when it detects that the key is pressed, key1=1, delay for a period of time and then detect again. If the key is still pressed, key1=2, and finally process according to the value of key1 in the main loop. This is what I did, and I can make one key play the role of two keys.