Swap String values without temporary variable

Question

I am aware of overflow, performance drops, etc but I need to swap two String values without any temporary variable. I know what cons are and found some effective methods but cannot understand one of them.

String a = "abc";
String b = "def";

b = a + (a = b).substring(0, 0);

System.out.printf("A: %s, B: %s", a, b);

Output shows it clear values are swapped. When I look at this seems something related with priority operations but I cannot figure it out in my mind. Please, could someone explain to me what happens?

What do you mean without using any temporary variable? Do you not want to see a temporary variable being used? Because I'm sure there'll always be a temporary variable that is being stored somewhere implicitely. — skiwi
– skiwi, Commented May 30, 2014 at 10:10
@skiwi It's implicitly, of course. At when dealing with immutable objects. — Chris
– Chris, Commented May 30, 2014 at 12:04

WoDoSc · Accepted Answer · 2014-05-30 09:28:18Z

7

Basically you can think on the swap operation

b = a + (a = b).substring(0, 0);

as

b = "abc" + (a = "def").substring(0, 0);

In this first step I simply substituted the variables with their values (except from the a in the parenthesis, since this is assigning a value, and not reading).

Now a is equal to "def" and therefore b is:

b = "abc" + "def".substring(0, 0);

In this step I substituted the parenthesis with its value. And now that is clearly:

b = "abc";

since the method .substring(0, 0) returns an empty string.

So the swap is done.

edited May 30, 2014 at 9:28

answered May 30, 2014 at 9:22

WoDoSc

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3 Comments

Thilo Over a year ago

That is probably not even that inefficient, as the only temporary string that is being created is of length zero. (I assume that appending a zero-length String just returns the input).

Thilo Over a year ago

"assume that appending a zero-length String just returns the input". Okay, probably not, as the process seems to have to involve a StringBuilder which has to copy things around.

Chris Over a year ago

Well, this is the answer I was looking for. Thank you @EnneCH for the explanation. I know this code is not usual coding style and it's not good practice but interviewers asks sometimes for such things.

Norbert Radyk · Accepted Answer · 2014-05-30 09:32:20Z

The fact you need to ask the question what's happening here, means you should seriously consider using constructions like this in your code.

Since in Java expressions are evaluated left to right (see here - 'Order of Evaluation' section).

This is what happens step-by-step in case of b = a + (a = b).substring(0, 0)

Populate a variable with its value:

b = "abc" + (a = b).substring(0, 0);
Populate b variable with its value:

b = "abc" + (a = "def").substring(0, 0);
Since first attribute of string concatenation (+) is already evaluated evaluate the second one, which starts with reassigning "def" to a.

b = "abc" + "def".substring(0, 0); //Now a = "def"
Run substring on "def"

b = "abc" + ""
Concatenate

b = "abc"
Reassign to b and now b = "abc" (while as we've shown before a has been reassigned to "def".

HGMamaci · Accepted Answer · 2017-08-30 14:09:33Z

0

Assuming both a and b are string variables:

a = (a+b).Substring((b=a).Length);

answered Aug 30, 2017 at 14:09

HGMamaci

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Comments

Vishnu Mari · Accepted Answer · 2023-07-10 11:15:51Z

0

public class swapping { public static void main(String are[]) {

    //Swapping two Integer without temp variable 
    int a=10;
    int b=20;
    a=a+b;
    b=a-b;
    a=a-b;
    System.out.println(a+" "+b);
    
    //Swapping two string character with temp variable.
    String str1="Good";
    int as=str1.length();
    String str2="Morning";
    str1=str1+str2;
    str2=str1.substring(0, as);
    str1=str1.substring(as);
    System.out.println(str1+" "+str2);
}

}

answered Jul 10, 2023 at 11:15

Vishnu Mari

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