Java中的参数传递及此情景下的String
2018-08-03 本文已影响0人
帕博雷克斯丢丢
public class Demo {
public static void main(String[] args) {
StrStb strb = new StrStb();
strb.add(strb.strCon);
System.out.println(strb.strCon);
strb.add(strb.strBuf);
System.out.println(strb.strBuf);
}
}
class StrStb {
public String strCon = "hello";
public String strObj = new String("dont BB");
public StringBuffer strBuf = new StringBuffer("string buffer");
public void add(String s) {
System.out.println(s == strCon);
System.out.println("1:" + s);
s = "No Hello";
System.out.println(s == strCon);
System.out.println("2:" + s);
System.out.println("3:" + strCon);
}
public void add(StringBuffer sb) {
sb.append("------FinalEnd");
}
}