Changing the Parameter Variable + Appending in String

[Dylan Kwon]

class X {
static void m (StringBuilder sb1) {
sb1.append(“er“);
}
public static void main(String[] args) {
StringBuilder sb2 = new StringBuilder(“moth“);
m(sb2);
System.out.println(sb2);
}
}

Output: mother
For the above code, why does changing sb1, the parameter variable, change sb2?

public class Hemlock {
static StringBuffer sb;
StringBuffer sb2;
public static void main(String[] args) {
sb.append(new Hemlock().go(new StringBuffer(“hey“)));
System.out.println(sb);
}
{
sb2 = new StringBuffer(“hi “);
}
StringBuffer go(StringBuffer s) {
System.out.print(s + “ oh “ + sb2);
return new StringBuffer(“ey“);
}
static { sb = new StringBuffer(“yo “);}
}

Output: hey oh hi yo ey
For the above code, why is "hey oh hi" appended before "yo?"

[Paul Clapham]

Dylan Kwon wrote:For the above code, why does changing sb1, the parameter variable, change sb2?

The code doesn't change the variable. It changes the StringBuilder which the variable refers to, which is also the StringBuilder which sb2 refers to. To change the variable would require code like "sb1 = ...".

For the above code, why is "hey oh hi" appended before "yo?"

It isn't exactly "appended" -- that isn't the word I would use. It's just written to the console before "yo" is written to the console. Look at the two System.out.println() statements and see what they print.

[Paul Clapham]

It's also important to know that StringBuilder's "append" method actually changes the contents of the StringBuilder object. Whereas the "append" method of String does NOT change the contents of the String object. If the question had used String instead of StringBuilder, the result would have been different.

[Dylan Kwon]

Thank you so much, Paul this cleared up every question I had and more!