public static void main(String[] args) {
StringBuffer A = new StringBuffer("a");
StringBuffer B = new StringBuffer("b");
oper(A, B);
System.out.println(A);
System.out.println(B);
}
public static void oper(StringBuffer A, StringBuffer B) {
A.append(B);
B = A;
}
请问为什么打印出来A = ab;而B = b?
StringBuffer A = new StringBuffer("a");
StringBuffer B = new StringBuffer("b");
oper(A, B);
System.out.println(A);
System.out.println(B);
}
public static void oper(StringBuffer A, StringBuffer B) {
A.append(B);
B = A;
}
请问为什么打印出来A = ab;而B = b?
A.append(B);操作后,影响了实参,所一打印的结果是AB.而B = A;不会影响到实参。
String A = new String("a");
String B = new String("b");
oper(A, B);
System.out.println(A);
System.out.println(B);
}
public static void oper(String A, String B) {
A.replace("a", "c");
B = A;
}
我现在是对A进行了操作,但是结果是a , b
有点蒙...
StringBuffer A
" A.append(B);操作后,影响了实参,所一打印的结果是AB "那么String A
A.replace("a", "c");方法就不会影响实参吗?