为什么在&，&，||的情况下输出不同? [英] Why is the output different in case of &&, &, ||?

问题描述

这是代码段

您能解释一下输出为何变化的原因吗

Can you explain why outputs are varying

1)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false, b;
        b = (t && ((i++) == 0));
        b = (f && ((i+=2) > 0));
        System.out.println(i);      
    }
}

在这种情况下的输出为1

output in this case is 1

2)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false, b;
        b = (t & ((i++) == 0));
        b = (f & ((i+=2) > 0));
        System.out.println(i);      
    }
}

在这种情况下的输出为3

output in this case is 3

3)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false, b;
        b = (t || ((i++) == 0));
        b = (f || ((i+=2) > 0));
        System.out.println(i);      
    }
}

在这种情况下的输出为2

output in this case is 2

4)

public static ShortCkt {
    public static void main(String args[]) {
        int i = 0;
        boolean t = true;
        boolean f = false, b;
        b = (t | ((i++) == 0));
        b = (f | ((i+=2) > 0));
        System.out.println(i);      
    }
}

在这种情况下的输出为3

output in this case is 3

推荐答案

为什么在&&，&，||情况下输出不同? ?

Why is the output different in case of &&, &, || ?

就像在C/C ++中一样，&&被懒惰"地求值，而&则不被求值.

Just as in C/C++ && is evaluated "lazily" while & is not.

如果a为false，则a && b将返回false，甚至不评估b.

If a is false then a && b will return false without even evaluating b.

a || b也是如此:如果第一个操作数a为true，则整个表达式为true，而第二个操作数b则从不求值.对于a | b，将同时评估a和b.

Same goes for a || b: If the first operand, a is true, the whole expression is true and the second operand, b is never evaluated. For a | b however, both a and b will be evaluated.

如果使用&&(或||)时未求值的操作数具有副作用，则会产生后果.

This has consequences if the operand that's not being evaluated when using && (or ||) has side effects, as in your examples.

侧面说明:很少有Java程序员知道^(xor)也适用于布尔值. (^^版本不仅仅因为它是多余的就不存在.)

Side note: Few java-programmers know that ^ (xor) works for booleans as well. (A ^^ version does not exist simply because it would be redundant.)

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