Question

所以我试图让时间功能倒计时。它基于我在这里看到的东西。变量由不同的函数给出。一旦时间用完，变量finish就会变为1并且它会离开函数。这个功能有时有效，有时它不起作用，例如，如果我给它一个11秒的输入它工作正常，但如果我给它1分钟它不起作用。有人能告诉我代码有什么问题。

if (time1 == 0 && time2 == 0 && time3 == 0 && time4 == 0)
//if all the time is 0 finish the sequence
    finish = 1;

if (time1 != 0) //Checking to see if the first digit is NOT at 0
    time1 = time1 - 1; //  subtract time 1 by 1
else {
    time2 = time2 - 1;  //When time1 is 0
    time1 = 9;
} //Time1 going back to it's original value

if (time2 == 0 && time1 == 0) { //if time1 and time2 are 0s
    if (time3 != 0) { //The minute value (time3)
        time2 = 5;  //60 SECONDS
        time3 = time3 - 1;
        time1 = 9;
    }             
} //Put time 1 to its original value
if (time2 <= 0 && time1 <= 0 && time3 <= 0) {
    if (time4 != 0) { //The minute value (time3)
        time2 = 5; //60 SECONDS
        time3 = 9;
        time4 = time4 - 1;
        time1 = 9; 
    }
} //Put time 1 to its original value

Time4 = 3，Time3 = 2，Time2 = 1，Time1 = 0.这意味着时间是32:10 min

Answer 1

您不能只检查非零，您需要检查给定时间是否为正，否则您将受到负值的计数，并且计数器可能会溢出。

if (time1 > 0)
    time1 -= 1;
if (time3 > 0)
    time3 -= 1;

另一个想法是，你用分钟和秒的每个数字倒计时，为什么不只是使用秒，将你的时间转换为秒。例如，倒计时1:23：

int minutes = 1;
int seconds = 23;
int timer = minutes * 60 + seconds;

// in your timer function
if (seconds == 0) {
    finish = 1;
} else if (seconds > 0) {
    seconds -= 1;
} else {
      // error
}

这样它也是可扩展的，如果你想处理几个小时，只需将hours * 3600添加到seconds，你就可以轻松地处理几天甚至几个月。在你的方法中，添加那些会导致太多情况，它们几乎不可能正确处理。

Answer 2

问题在于，在将数字更改为非零后，您将比较为零。

假设1:00编码为

time1 = 0
time2 = 0
time3 = 1

你可以遵循自己的逻辑：

if (time1 != 0) // Nope
    time1 = time1 - 1;
else { // Yes
    time2 = time2 - 1;
    time1 = 9;
}

现在你有了

time1 == 9
time2 == 0
time3 == 1

if (time2 == 0 && time1 == 0) { // Nope, time1 is 9
    if (time3 != 0) {
        time2 = 5;  
        time3 = time3 - 1;
        time1 = 9;
    }             
}

你还有

time1 == 9
time2 == 0
time3 == 1

最后

if (time2 <= 0 && time1 <= 0 && time3 <= 0) { // Nope
    if (time4 != 0) { 
        time2 = 5; 
        time3 = 9;
        time4 = time4 - 1;
        time1 = 9; 
    }
}

所以你最终得到了

time1 == 9
time2 == 0
time3 == 1

即1:09。

您想要更改时间_k的唯一时间是_k-1时间“越过”零时。
这可以通过一组条件来完成：

if (time1 > 0 || time2 > 0 || time3 > 0 || time4 > 0)
{    
    time1 -= 1;
    if (time1 < 0)
    {
        time1 = 9;
        time2 -= 1;
        if (time2 < 0)
        {
            time2 = 5;
            time3 -= 1;
            if (time3 < 0)
            {
                // ...
            }
        }
    }
}

时间函数有时有时无效

2 个答案: