Java String array created inside loop

Question

I have a question regarding the differences of the following codes:

Vector v = new Vector();
String [] str_arr = new String[3];

for(int i=0; i<3; i++)
{
   str_arr[0] = "A";
   str_arr[1] = "B";
   str_arr[2] = "C";            
   v.add(str_arr);
}
System.out.println(v.size());  //answer 3

Versus

Vector v = new Vector();


for(int i=0; i<3; i++)
{
   String [] str_arr = new String[3];
   str_arr[0] = "A";
   str_arr[1] = "B";
   str_arr[2] = "C";            
   v.add(str_arr);
}
System.out.println(v.size());  //answer 3

The only difference between both codes is, for the second one, string array is created inside the loop.

Both the codes produce same result, but i want to know what is the difference between these two.

In the second code snippet you creating the array 3 times - that means, the performance is lower — i23
– i23, Commented Jan 7, 2016 at 13:59
Don't use Vector, it isn't useful in 99% cases. It's a deprecated collection — coolguy
– coolguy, Commented Jan 7, 2016 at 14:04
@coolguy Except that it's not deprecated, and it's faster than a synchronized ArrayList. — John E.
– John E., Commented Jan 7, 2016 at 14:07
@coolguy which collection you suggest me to use? im using java 1.4.2 — hades
– hades, Commented Jan 7, 2016 at 14:09
@JohnE. I think you have it backwards. Vector is synchronized (at least since Java 1.5). ArrayList is not. You are correct that is not deprecated, but its use is discouraged when thread safety is not an issue. — Erick G. Hagstrom
– Erick G. Hagstrom, Commented Jan 7, 2016 at 14:18

Eran · Accepted Answer · 2016-01-07 14:00:07Z

7

The two snippets don't produce the same result. The first snippet adds the same array object 3 times to the Vector. The second snippet adds three different array objects to the Vector.

The results may seem the same, since all three arrays in the second snippet contain the same values.

If you'd change the assignment from

str_arr[0] = "A";
str_arr[1] = "B";
str_arr[2] = "C";

to

str_arr[0] = "A" + i;
str_arr[1] = "B" + i;
str_arr[2] = "C" + i;

You'd see that in the first snippet, the all the arrays in the Vector contain [A2,B2,C2], since there's just one array that's getting overwritten.

On the other hand, the second snippet would produce a Vector that contains three different arrays - [A0,B0,C0],[A1,B1,C1],[A2,B2,C2].

edited Jan 7, 2016 at 14:00

answered Jan 7, 2016 at 13:58

Eran

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2 Comments

MrHug Over a year ago

Perhaps you can give a small example of how this result can be observed? I.e. getting the first element from v and modifying it, will also modify the second and third element in v for the first code fragment.

hades Over a year ago

@Eran thanks, sorry i was overlook, just looking at the size but not the content, thanks for the example.

Arc676 · Accepted Answer · 2016-01-07 14:04:43Z

2

In the second version, the String[] goes out of scope at the end of each iteration. This means that you can't access it elsewhere. After the loop, the variable no longer exists. You also don't keep the value between iterations.

EDIT: The array itself still exists, as the vector keeps a strong reference to it.

However, there are some redundancy issues in your code.

for(int i=0; i<3; i++)
{
   str_arr[0] = "A"; //all
   str_arr[1] = "B"; //of this
   str_arr[2] = "C"; //is redundant  
   v.add(str_arr);
}

Here (first version), you assign the same values to the array each iteration even though the variable still exists from one iteration to another, so you don't need most of the body

edited Jan 7, 2016 at 14:04

answered Jan 7, 2016 at 14:00

Arc676

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1 Comment

Arc676 Over a year ago

@ErickG.Hagstrom Thanks. Edited.

Andres · Accepted Answer · 2016-01-07 14:06:25Z

1

Though the output will be the same, the state of your application will be different at the last line of both programs.

On the first, your heap will have only one array of Strings, and on the second it will have three.

answered Jan 7, 2016 at 14:06

Andres

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Comments

user3538022 · Accepted Answer · 2016-01-07 14:16:43Z

0

The first snippet will be a little faster than second snippet (tack less Memory and process) because it implement the array once and update it's values in each loop but in the second snippet it's implement new array and register an address for it in each loop

answered Jan 7, 2016 at 14:16

user3538022

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