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Java——如何判断整数溢出


问题:

在之前刷题的时候遇见一个问题,需要解决int相加后怎么判断是否溢出,如果溢出就返回Integer.MAX_VALUE。

解决方案:

加法

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

public static int subtractExact(int x, int y) {
int r = x - y;
if (((x ^ y) & (x ^ r)) < 0) {
throw new ArithmeticException("integer overflow");
}
return r;
}

乘法

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;
}

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;
}

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