Java如何判断整数溢出，溢出后怎么得到提示

￡神魔★判官ぃ 2023-02-13 14:47 66阅读 0赞

# 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;
        }

# 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;
        }