Java Notes: Some Tricky Problems in Java
2017-02-04 06:17
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1/ About the modulus:
对于求余运算需要注意,当被除数是一个负数的时候,余数永远都是负的。所以你判断是否为奇数的条件不能是:x % 2 == 1,而是 x % 2 != 0。
2/ 如何提取出一个数的个、十...位(从低到高)
3/ Keep in mind of overflow problem
如果int到了最大的2^30次方,不能再乘以2(右移一位)了。强行再右移的话,会瞬间变为最小的int,值是-2^31次方。下面的程序就有这样的问题:
上述程序尝试测试一个数是否是2的倍数。然而,当所给的参数大于int的2^31 - 1时候,会产生overflow问题,程序进入死循环。
4/
对于求余运算需要注意,当被除数是一个负数的时候,余数永远都是负的。所以你判断是否为奇数的条件不能是:x % 2 == 1,而是 x % 2 != 0。
2/ 如何提取出一个数的个、十...位(从低到高)
public Class Example { public static void main(String[] args) { int x = Integer.parseInt(args[0]); while (x != 0) { int digit = Math.abs(x % 10); // 如果不加math.abs的话如果被余数是负的,结果都是负数来的 System.out.println(digit); x = x / 10; } } }
3/ Keep in mind of overflow problem
如果int到了最大的2^30次方,不能再乘以2(右移一位)了。强行再右移的话,会瞬间变为最小的int,值是-2^31次方。下面的程序就有这样的问题:
public class Example { public static void main(String[] args) { int x = Integer.parseInt(args[0]); boolean answer = false; int p = 1; while (p <= x) { System.out.println("Testing" + p); if (p == x) { answer = true; } p = p * 2; } System.out.println(answer); } }
上述程序尝试测试一个数是否是2的倍数。然而,当所给的参数大于int的2^31 - 1时候,会产生overflow问题,程序进入死循环。
4/
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