在C和C中,根据Shafik’s post,INT_MIN％-1的行为似乎是未定义/平台相关的.

在Java中,％运算符是否会溢出？

考虑一下这段代码：

public class Test {
    public static void main(String[] args) {
        // setup variables:
        byte b = Byte.MIN_VALUE % (-1);
        short s = Short.MIN_VALUE % (-1);
        int i = Integer.MIN_VALUE % (-1);
        long l = Long.MIN_VALUE % (-1);

        // my machine prints "0" for all:
        System.out.println(b);
        System.out.println(s);
        System.out.println(i);
        System.out.println(l);
    }
}

是否存在与平台无关的保证,上述结果为0？

解决方法:

看看JLS section 15.17.3它说：

In C and C++, the remainder operator accepts only integral operands,
but in the Java programming language, it also accepts floating-point
operands.

The remainder operation for operands that are integers after binary
numeric promotion (§5.6.2) produces a result value such that
(a/b)*b+(a%b) is equal to a. This identity holds even in the special
case that the dividend is the negative integer of largest possible
magnitude for its type and the divisor is -1 (the remainder is 0). It
follows from this rule that the result of the remainder operation can
be negative only if the dividend is negative, and can be positive only
if the dividend is positive;