这段代码将一个字符串分离为令牌，并将它们存储在一个字符串数组中，然后将一个变量与第一个home…为什么它不工作?
public static void main(String...aArguments) throws IOException {
String usuario = "Jorman";
String password = "14988611";
String strDatos = "Jorman 14988611";
StringTokenizer tokens = new StringTokenizer(strDatos, " ");
int nDatos = tokens.countTokens();
String[] datos = new String[nDatos];
int i = 0;
while (tokens.hasMoreTokens()) {
String str = tokens.nextToken();
datos[i] = str;
i++;
}
//System.out.println (usuario);
if ((datos[0] == usuario)) {
System.out.println("WORKING");
}
}
如果你要比较字符串的任何赋值，即原始字符串，"=="和.equals都可以，但对于新的字符串对象，你应该只使用.equals，在这里"=="将不起作用。
例子:
String a = "name";
String b = "name";
If (a == b) and (a.equals(b))将返回true。
But
String a = new String("a");
在这种情况下，if(a == b)将返回false
所以最好使用.equals操作符…
==测试引用是否相等。
.equals()测试值是否相等。
因此，如果你真的想测试两个字符串是否具有相同的值，你应该使用.equals()(除非在少数情况下，你可以保证具有相同值的两个字符串将由相同的对象表示，例如:字符串实习)。
==用于测试两个字符串是否是同一个Object。
// These two have the same value
new String("test").equals("test") ==> true
// ... but they are not the same object
new String("test") == "test" ==> false
// ... neither are these
new String("test") == new String("test") ==> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" ==> true
// concatenation of string literals happens at compile time resulting in same objects
"test" == "te" + "st" ==> true
// but .substring() is invoked at runtime, generating distinct objects
"test" == "!test".substring(1) ==> false
值得注意的是，==比equals()更便宜(一个指针比较而不是一个循环)，因此，在适用的情况下(例如，您可以保证您只处理实习字符串)，它可以提供重要的性能改进。然而，这种情况很少见。
It's good to notice that in some cases use of "==" operator can lead to the expected result, because the way how java handles strings - string literals are interned (see String.intern()) during compilation - so when you write for example "hello world" in two classes and compare those strings with "==" you could get result: true, which is expected according to specification; when you compare same strings (if they have same value) when the first one is string literal (ie. defined through "i am string literal") and second is constructed during runtime ie. with "new" keyword like new String("i am string literal"), the == (equality) operator returns false, because both of them are different instances of the String class.
唯一正确的方法是使用.equals() -> datos[0].equals(通常)。==表示仅当两个对象是object的相同实例(即。内存地址相同)
更新:01.04.2013我更新了这篇文章，因为下面的评论是正确的。最初我声明实习(String.intern)是JVM优化的副作用。虽然它确实节省了内存资源(这就是我所说的“优化”)，但它是语言的主要特征
