Java 逻辑运算符短路

本教程将介绍Java 逻辑运算符短路的处理方法，这篇教程是从别的地方看到的，然后加了一些国外程序员的疑问与解答，希望能对你有所帮助，好了，下面开始学习吧。

问题描述

Which set is short-circuiting, and what exactly does it mean that the complex conditional expression is short-circuiting?

public static void main(String[] args) {
  int x, y, z;

  x = 10;
  y = 20;
  z = 30;

  // T T
  // T F
  // F T
  // F F

  //SET A
  boolean a = (x < z) && (x == x);
  boolean b = (x < z) && (x == z);
  boolean c = (x == z) && (x < z);
  boolean d = (x == z) && (x > z);
  //SET B 
  boolean aa = (x < z) & (x == x);
  boolean bb = (x < z) & (x == z);
  boolean cc = (x == z) & (x < z);
  boolean dd = (x == z) & (x > z);

}

解决方案

The && and || operators "short-circuit", meaning they don't evaluate the right-hand side if it isn't necessary.

The & and | operators, when used as logical operators, always evaluate both sides.

There is only one case of short-circuiting for each operator, and they are:

false && ... - it is not necessary to know what the right-hand side is because the result can only be false regardless of the value there

true || ... - it is not necessary to know what the right-hand side is because the result can only be true regardless of the value there

Let's compare the behaviour in a simple example:

public boolean longerThan(String input, int length) {
 return input != null && input.length() > length;
}

public boolean longerThan(String input, int length) {
 return input != null & input.length() > length;
}

The 2nd version uses the non-short-circuiting operator & and will throw a NullPointerException if input is null, but the 1st version will return false without an exception.

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