java - Java 的 % 运算符会溢出吗？

Question

在 C 和 C++ 中，根据 Shafik's post，INT_MIN % -1 的行为似乎未定义/平台相关.

在 Java 中，% 运算符会溢出吗？

考虑这段代码:

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);
    }
}

是否有独立于平台的保证上述结果为0？

Answer 1

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;