if ((time1 - time2) < interval) { return true; } return false; } 将这段代码改成一个return
时间: 2024-02-14 22:14:47 浏览: 96
您可以使用条件运算符将这段代码改写成一个 return 语句,如下所示:
```
return ((time1 - time2) < interval) ? true : false;
```
这将首先计算表达式 `(time1 - time2) < interval` 的值,如果为 true,则返回 true,否则返回 false。这种写法可以使代码更加简洁,但有些人认为可读性稍差。
相关问题
public static boolean validateCode(String secret, int code, long time) throws NoSuchAlgorithmException, InvalidKeyException { for (int i = -WINDOW_SIZE; i <= WINDOW_SIZE; i++) { long t = time + i * INTERVAL; int c = generateCode(secret, t); if (c == code) { return true; } } return false; } 我想增加一个参数 ss ss可不填,不填时INTERVAL取默认值,当ss传入参数且为int时INTERVAL=ss,当ss传入参数但不为int时,抛出错误
可以在方法的参数列表中增加一个参数 ss,并在方法中加入如下代码:
```
int interval = INTERVAL;
if (ss != null) {
if (ss instanceof Integer) {
interval = (int) ss;
} else {
throw new IllegalArgumentException("ss must be an integer");
}
}
```
这段代码的作用是:如果 ss 不为 null,首先判断它是否为 Integer 类型,如果是,则将其转换为 int 类型并赋值给 interval 变量;如果不是,则抛出 IllegalArgumentException 异常。如果 ss 为 null,则不对 interval 进行修改,仍然取默认值 INTERVAL。
接下来,在调用 generateCode 方法时,将 interval 作为参数传入即可:
```
int c = generateCode(secret, t, interval);
```
这样就完成了对方法的修改。
if 0 != iteration: threading.Timer(delay, repeat, (fn, iteration - 1, interval, True)).start()优化这段代码
One way to optimize this code is to use the built-in `time.sleep()` function instead of creating a new thread for each iteration. This can be done by modifying the `repeat()` function as follows:
```
import time
def repeat(fn, iteration, interval, threaded=False):
if iteration == 0:
return
fn()
time.sleep(interval)
if threaded:
threading.Timer(interval, repeat, (fn, iteration - 1, interval, True)).start()
else:
repeat(fn, iteration - 1, interval)
```
This modified function will use the `time.sleep()` function to pause the current thread for the specified interval, instead of creating a new thread. This can lead to better performance and reduced overhead. We also added a `threaded` parameter to allow the function to be used in both threaded and non-threaded contexts.
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