Objects in String and StringBuffer

[suhaas_mohandos]

How many objects are created in Code 1 in each line and also in Code 2 in each line? How is StringBuffer better when compared to String?

Code 1:

1. String output = “Some text”
2. Int count = 100;
3. for(int i =0; i<count; i++) {
4. output += i;
5. }
6. return output;

Code 2:

1. StringBuffer output = new StringBuffer(110);// set an initial size of 110
2. output.append(“Some text”);
3. for(int i =0; i<count; i++) {
4. output.append(i);
5. }
6. return output.toString();

[JosAH]

In line #4 of the first example the following objects are created:

1) a temporary StringBuilder (or StringBuffer)
2) a String representation of variable 'i'
3) a String representation of the StringBuilder.

In line #4 of the second example, only a String representation of variable 'i' is created.

kind regards,

Jos

[Tolls]

What do you think is the answer?
Otherwise we might just think you're dumping your homework here...

Besides, I hate these things...I mean, we don't know how many objects are created for some of those lines as the constructor of (eg) the StringBuffer may be (and I expect is) creating objects for its attributes.

Or am I overthinking things again?

[masijade]

Or am I overthinking things again?

Probably, as these questions (whether homework or exam) are usually only worried about the "main" objects, i.e. the StringBuffer and String object creations. They never care to know about the char arrays, and ints, for example, that are part of the String and StringBuffer objects.

[gimbal2]

Looks like the homework question here is to actually explain why StringBuffer/StringBuilder may actually be a slightly better choice when building a larger String; it isn't so much about the exact number of objects created, just to realize and explain that listing one will in fact create many more objects than listing 2.

I'm not going to give the answer to that, it would defeat the purpose of the homework question. Try to find a possible explanation first and then we can discuss how much of it is true.

[jim829]

Is it a reasonable requirement for some to have to know how the compiler generates byte codes for various constructs? It used to be that compilers from different vendors would generate different code. It would all execute but some generated code would be better than others. And no one could be expected to know from architecture to architecture what the generated machine code would look like.

But even with a virtual architecture there could be differences in generated byte code between different vendor implementations. Or have the Java specifications eliminated that and dictated how code should be compiled?

Regards,
Jim

[masijade]

The actual byte code can still vary, somewhat, but the behaviour is fairly strictly defined, so the differences SHOULD be VERY minor.

[JosAH]

The only thing the JLS has to say about it is this:

15.18.1.2 Optimization of String Concatenation

An implementation may choose to perform conversion and concatenation in one step to avoid creating and then discarding an intermediate String object. To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.
For primitive types, an implementation may also optimize away the creation of a wrapper object by converting directly from a primitive type to a string.

kind regards,

Jos

[jim829]

The reason I ask is that this is the second time in a week that a question has come up regarding how the code is compiled. The first was why one uses int and not short or byte in for loops. I had to write some examples and look at the compiled byte codes to realize that there are special instructions for handling int types that make it more efficient. Had I not done this I would not have known the answer.

I used to program in Digital Equipment Corporations PDP/11 assembly language. It had identical instructions for both bytes and ints. So no performance improvement would necessarily be gained using one over the other. And I would never have been asked a similar question about a C compiler because it depended on the architecture. So it seems that one must know much more about the internals of Java to be an expert Java programmer than would be required to be an expert C (or its variants) programmer.

Regards,
Jim

[gimbal2]

The reason I ask is that this is the second time in a week that a question has come up regarding how the code is compiled. The first was why one uses int and not short or byte in for loops. I had to write some examples and look at the compiled byte codes to realize that there are special instructions for handling int types that make it more efficient. Had I not done this I would not have known the answer.

Okay, but you won't notice that difference unless you're doing some massively CPU heavy code. Better not to worry about such internal details until you need to. And by extension: better not make novices aware of such internal details, because there is already too much of a tendency to want to do premature optimizations.

[jim829]

I agree entirely. I just wanted to see if I could determine why the int based for loop was preferable. Remember -- this was from an earlier post and was supposedly asked as an interview question (which I personally believe is an unreasonable question). If there is another, less anal answer to the question I would like to know. I would have probably answered it wrong.

Regards,
Jim

[gimbal2]

IMO - it is unreasonable to ask it in general. In very specific areas of interest, namely gaming & simulation and real time processing or other performance intensive applications, then it surely helps a whole lot to know how Java ticks under the hood. As an extreme example: I've seen people tuning their application such that a garbage collection was only triggered once at night in a window in which the application was allowed to be unresponsive for an hour. You can't get that done unless you dig in deep.

But in the general case you have to write dumb code and let the Java compiler and runtime optimizer do its thing. Not making use of the Java platform features basically defeats the purpose of using Java after all. Forgetting about byte and short and always using int is nice and dumb in my book :)

Long story short: nowadays I would actually encourage to not know too much about it; the less you know, the less chance you'll have of throwing a wrench into the machine that is Java.

[suhaas_mohandos]

This is not a homework question.

Anyways my take on this

Code 1:

String output = “Some text”
Int count = 100;
for(int i =0; i<count; i++) {
output += i;
}
return output;

In code 1:
in line 1, one object is created
in line 4, hundred objects are created(from 0-99)
Total:101 objects are created

Code 2:

StringBuffer output = new StringBuffer(110);// set an initial size of 110
output.append(“Some text”);
for(int i =0; i<count; i++) {
output.append(i);
}
return output.toString();

In code 2:
in line 1 one object is created
Total: 1 object is created.

Pls correct me if I am wrong

BTW is there a difference between int and Int?

[Junky]

In code 1:
in line 1, one object is created

I maybe wrong but I believe it is 2. "some text" is created and placed in the String Literal Pool and a String object which references that literal is created and placed in the normal object heap.

BTW is there a difference between int and Int?

Try compiling the second one.

[gimbal2]

This is not a homework question.

It may not be homework, but it is certainly a homework question. Lets replace homework with "quiz" : nothing changes, the purpose is that you think about it and provide your own vision. Judging by what you've given now, I see you've got the right idea.

[JosAH]

In code 1:
in line 1, one object is created
in line 4, hundred objects are created(from 0-99)
Total:101 objects are created

That is not correct; see my reply #2.

kind regards,

Jos

[Tolls]

I maybe wrong but I believe it is 2. "some text" is created and placed in the String Literal Pool and a String object which references that literal is created and placed in the normal object heap.

Nope.
No objects are created on that line.
A reference to the object in the String Literal Pool containing "some text" is assigned to that String variable.
The pool already exists.