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");
}
}
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
The == operator checks if the two references point to the same object or not.
.equals() checks for the actual string content (value).
注意.equals()方法属于类Object(所有类的超类)。您需要根据您的类需求重写它,但是对于String,它已经实现,并且它检查两个字符串是否具有相同的值。
Case1)
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s1; // true
s1.equals(s2); // true
Reason: String literals created without null are stored in the string pool in the permgen area of the heap. So both s1 and s2 point to the same object in the pool.
Case2)
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; // false
s1.equals(s2); // true
Reason: If you create a String object using the `new` keyword a separate space is allocated to it on the heap.