计数器的C程序

Question

我想创建一个计数器。每隔 10 秒我想检查一次开关的状态。例如，如果开关闭合，则 10 秒计数器递增。如果打开，它将返回 sleep 状态，10 秒后再次唤醒并检查开关的状态。当计数达到例如 100 时，就做一些事情。我该怎么做

我的尝试是:

for(int i=0;i<100;i++){
if(SW=1) {
    i++;
}
else
    i=0;
}

Answer 1

我认为你需要更具体地回答这个问题。似乎您想在每次打开开关时重置计数器。您确定要这样吗？

无论如何，这可能就是你想要的

#include <stdio.h>
#include <time.h>
struct tm tm;   
time_t start,enxd;
double sec;
int counter;
int main(){
    int switchCounter = 0;
    int checkSwitch;

    checkSwitch = 1; // I put this in purpose since I have no idea how you're going to read the switch. 
                     // Thus, this assumes the switch is always closed.

    while(switchCounter != 100){
        // 1. Wait for 10 seconds
        sec = 0;
        time(&start);

        while(sec !=10){
            ++counter;
            time(&enxd);
            sec=difftime(enxd,start);
        }

        // 2. Read the state of the switch here.
        // ..............

        // 3. Simple if-else
        if (checkSwitch == 1){ //switch is closed
            switchCounter++;
            printf("Counter incremented. Current = %i \n", switchCounter);
        }
        else //if switch is open
        {
            switchCounter = 0 ;// Iam confused here, is this what you want ?
            printf("Switch is open \n");
        }
    }
    // 4. Finally when your counter reaches 100, you wanna do something here
    // ............................

    return 0;
}

希望有帮助:)