compare two numeric String values

Question

I have a simple doubt. Would be great if anyone helps me.

I have two strings:

String string1 = "4"; // and 
String string2 = "04";

Both the values are equal but how to compare them in java? We have equals and equalsIgnoreCase for comparing String alpha values, similarly how to compare numeric values.

@madhairsilence. Note that your method will fail, as you are comparing two Integer objects using ==, which might be considered a Crime. — Rohit Jain
– Rohit Jain, Commented Apr 2, 2013 at 12:39
@RohitJain please note that integers created by new are not 'cached'. Integer a = 1; is not the same as Integer a = new Integer(1); — msi
– msi, Commented Apr 2, 2013 at 12:41
@madhairsilence. 2. Java Caches the Integer objects in a certain range, which is generally [-128, 127], in which case, it will not create a new Integer object when you assign an integer literal to an Integer reference, as stated by msi. I was mistaken, Cache doesn't hold tru in your case. — Rohit Jain
– Rohit Jain, Commented Apr 2, 2013 at 12:51
@madhairsilence. 3. Crime with object comparison with == is that, it doesn't really compares the content, rather the value of reference. So, two object reference pointing to two objects having same value will not be judged equal by ==, so you should always use equals method for object comparison. — Rohit Jain
– Rohit Jain, Commented Apr 2, 2013 at 12:53

Rohit Jain · Accepted Answer · 2013-04-02 12:36:08Z

24

Integer.parseInt("4") == Integer.parseInt("04")

That is it. You can convert a numeric string into integer using Integer.parseInt(String) method, which returns an int type. And then comparison is same as 4 == 4.

answered Apr 2, 2013 at 12:36

Rohit Jain

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1 Comment

Harshil Lodhi Over a year ago

Wouldn't this overflow if the numbers are long than what Int can hold

marstran · Accepted Answer · 2017-06-28 11:52:09Z

17

Don't forget the BigInteger for very long values.

return new BigInteger(s1).compareTo(new BigInteger(s2));

edited Jun 28, 2017 at 11:52

marstran

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answered Dec 12, 2015 at 10:55

Jin Kwon

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4 Comments

Muflix Over a year ago

BigInteger does not accept string.

Jin Kwon Over a year ago

@Muflix Yes it does. docs.oracle.com/en/java/javase/21/docs/api/java.base/java/math/…

Muflix Over a year ago

My apology, I was looking into .NET instead of Java 😹

Jin Kwon Over a year ago

@Muflix No problem. Have a nice day.

Deepak Bala · Accepted Answer · 2013-04-02 12:35:20Z

9

Integer.valueOf(string1).equals(Integer.valueOf(string2));

answered Apr 2, 2013 at 12:35

Deepak Bala

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1 Comment

Blank Chisui Over a year ago

valueOf(String) internally calls parseInt(String) and wraps it. Comparing the results of parseInt(String) using == works without boxing.

Viktor Mellgren · Accepted Answer · 2017-06-28 12:06:53Z

4

Skipping parsing it to an int at all to handle arbitrarily large numbers, only comparing the strings. As a bonus it also works for decimal numbers (unless there are trailing zeroes, where you probably can do similar thing with anchoring end of string instead).

You can omitt the trim if you are sure there are no whitespace characters

    String string1 = "4"; // and
    String string2 = "04";

    String noLeadingZeroes1 = string1.trim().replaceFirst("^0+(?!$)", "");
    String noLeadingZeroes2 = string2.trim().replaceFirst("^0+(?!$)", "");

    System.out.println(noLeadingZeroes1.equals(noLeadingZeroes2));

answered Jun 28, 2017 at 12:06

Viktor Mellgren

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1 Comment

Al-Alamin Over a year ago

I also found this answer's explanation very helpful. stackoverflow.com/a/2800839/3966480 The ^ anchor will make sure that the 0+ being matched is at the beginning of the input. The (?!$) negative lookahead ensures that not the entire string will be matched.

Afshin Moazami · Accepted Answer · 2013-11-25 05:11:44Z

3

Use the Integer class to convert the strings to integers.

http://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#parseInt%28java.lang.String%29

Integer.parseInt(string1) == Integer.parseInt(string2)

edited Nov 25, 2013 at 5:11

Afshin Moazami

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answered Apr 2, 2013 at 12:36

Colin D

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Comments

yakob abada · Accepted Answer · 2021-06-04 15:31:59Z

1

The comparison happens on chacracter basis. so '2' > '12' is true because the comparison will happen as '2' > '1' and in alphabetical way '2' is always greater than '1' as unicode. SO it will comeout true.

answered Jun 4, 2021 at 15:31

yakob abada

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Comments

Animesh Nag · Accepted Answer · 2022-06-19 21:28:43Z

-1

if(String.valueOf(S1).compareTo(String.valueOf(S2)) < 0)

answered Jun 19, 2022 at 21:28

Animesh Nag

9

1 Comment

Community Over a year ago

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