Java - Why is s not changed?

Umekichi · 09-01-2014 3:14pm #1

Hi there,
Currently doing some enthuware practice questions and came across this:

public class Sample{
public static void main(String[] args) {
String s1 = new String("java");
StringBuilder s2 = new StringBuilder("java");
replaceString(s1);
replaceStringBuilder(s2);
System.out.println(s1 + s2);
}
static void replaceString(String s) {
s = s.replace('j', 'l');
}
static void replaceStringBuilder(StringBuilder s) {
s.append("c");
}
}

ok, so I chose the answer lavajavac but the answer actually is javajavac.
Why is this?
Surely in replaceString, as it takes a String and assigns that String to the value of lava, then it would mean that the answer would be lavajavac?
Is this down to there being no "this" keyword?

Thanks

stevenmu · 09-01-2014 3:50pm

I'm guessing this question is about the differences between passing parameters by value and by reference.

When you pass by value, you are effectively making a copy of that value and sending that to the method, if you update that value in your method, it does not update the original value. When you pass by reference, you are passing a pointer to the original variable, if you update that value in your method, you are in fact updating the original variable.

IIRC, in Java, by default 'simple' types are passed by value and 'complex' types are passed by reference. Simple types are the built in basic types like int, float etc. Complex types are 'objects'. String is, I believe a simple type, and so should be passed by value, and StringBuilder is a complex type and should be passed by reference.

Following that logic though, the output should be "javajavac". I don't know where "javalavac" comes from.

Umekichi · 09-01-2014 3:52pm

stevenmu wrote: »

Following that logic though, the output should be "javajavac". I don't know where "javalavac" comes from.

Sorry typo, my eyes are going square from all this code

Also Thanks, I understand it a bit better

dsrckiyisvddht · 09-01-2014 4:18pm

Reason for this is that String is immutable.

In the below you are not returning anything and assigning it back to s1 so s1 remains the same.

static void replaceString(String s) {
 s = s.replace('j', 'l');
 }

If you change that to

static void replaceString(String s) {
 s = s.replace('j', 'l');
System.out.println(s);
 }

you will see it is only "lava" for the scope of that method

croo · 09-01-2014 7:37pm

stevenmu wrote: »

Following that logic though, the output should be "javajavac". I don't know where "javalavac" comes from.

and that IS the output!

chrislad · 10-01-2014 11:50am

If you wanted, you could also return a String

static String replaceString(String s) {
return s.replace('j', 'l');
}

So for String s1, you would have

s1 = replaceString(s1);

Java - Why is s not changed?

Comments