整数溢出
int类型一般占4个字节，故取值范围 -2^31 ~ 2^31-1,对于无符号另当别论，我们的讨论建立在以补码形式存储的带符号整数
-上溢
存储数值超过了整形数值2^31-1，导致数据向上溢出
-下溢
整数数值小于了-2^31，导致数值向下溢出
加法溢出
public static int addExact(int x, int y) {
int r = x + y;
// HD 2-12 Overflow iff both arguments have the opposite sign of the result
if (((x ^ r) & (y ^ r)) < 0) {
throw new ArithmeticException("integer overflow");
}
return r;
}
另一种写法：
if（num1>Integer.MAX_VALUE-num2)
System.out.println("num1+num2上溢出");
if（num1 System.out.println("num1+num2下溢出"); 减法溢出 public static int subtractExact(int x, int y) { int r = x - y; // HD 2-12 Overflow iff the arguments have different signs and // the sign of the result is different than the sign of x if (((x ^ y) & (x ^ r)) < 0) { throw new ArithmeticException("integer overflow"); } return r; } 原理其实很简单。只有两个正数或者两个负数相加的时候才会溢出，一正一负是不会溢出的。并且，溢出后得到的结果一定是与原值的符号相反的。 借用这个原理，(x ^ r) & (y ^ r)将用运算符算出来的r与x和y分别异或。异或的规则是同0异1。我们只看符号位，如果符号位不同，那么两个括号得到的都是1。而与运算又是只有1 & 1才是1，有一个为0都是0。因此，一旦溢出，无论是正溢出负溢出，(x ^ r) & (y ^ r)计算出的结果都应该小于0 public static void main(String[] args) { int c = Integer.MAX_VALUE + Integer.MAX_VALUE; long c1 = Integer.MIN_VALUE + Integer.MIN_VALUE; System.out.println(c); //-2 = 两正数相加，溢出最大到-2 System.out.println(c1); //0 = 两负数相加，溢出最大到0 // 两者符号不同 } 乘法溢出 public static int multiplyExact(int x, int y) { long r = (long)x * (long)y; if ((int)r != r) { throw new ArithmeticException("integer overflow"); } return (int)r; } 注意这个坑 public static void main(String[] args) { int a = 964632435; int c = a * 10; long c1 = a * 10; // 相当于(long)(a + 10) = (long)c：两个int类型运算，结果仍然是int，当 = 时，自动提升类型 long c2 = (long) (a * 10); long c3 = (long) c; System.out.println(c); //1056389758 System.out.println(c1); //1056389758 System.out.println(c2); //1056389758 System.out.println(c3); //1056389758 System.out.println(); long c4 = (long)a * (long)10; long c5 = (long)a *10; // int与long运算，int自动类型提升 System.out.println(c4); //9646324350 System.out.println(c5); //9646324350 System.out.println((int)c5); } 注意 long和int是不一样的 public static long multiplyExact(long x, long y) { long r = x * y; long ax = Math.abs(x); long ay = Math.abs(y); if (((ax | ay) >>> 31 != 0)) { // Some bits greater than 2^31 that might cause overflow // Check the result using the divide operator // and check for the special case of Long.MIN_VALUE * -1 if (((y != 0) && (r / y != x)) || (x == Long.MIN_VALUE && y == -1)) { throw new ArithmeticException("long overflow"); } } return r; }