java - 逻辑或和短路或运算符有什么区别？

Question

I am getting the out same output: a | b = false a || b = false a | b = true a || b = true a | b = true a || b = true a | b = true a || b = true

What is the difference between these two operators ? 

boolean a = false;  
boolean b = false;  

System.out.println("a | b = " + (a|b) );
System.out.println("a || b = " + (a||b) );

a = false;  
b = true;   
System.out.println("a | b = " + (a|b) );
System.out.println("a || b = " + (a||b) );

a = true;   
b = false;  
System.out.println("a | b = " + (a|b) );
System.out.println("a || b = " + (a||b) );

a = true;   
b = true;   
System.out.println("a | b = " + (a|b) );
System.out.println("a || b = " + (a||b) );

Answer 1

区别在于，如果第一个操作数为 true，短路运算符不会计算第二个操作数，而没有短路的逻辑 OR 始终计算两个操作数。

在简单的测试中你不会看到任何差异，因为假设没有抛出异常，两者应该给出相同的输出，但如果你尝试这样的事情:

String s = null;
System.out.println("a || b = " + s==null || s.length() == 0 );
System.out.println("a | b = " + s==null | s.length() == 0 );

第一个运算符将为您提供 true，而第二个运算符将为您提供 NullPointerException ，因为只有 |运算符将尝试评估 s.length() == 0 .