NumberFormatException while converting a binary string to a number

Question

So I have this binary string (32 chars corresponding to 32 bits):

 String s = "11111111110000011101110111011000";

When I try to convert it to a number, it throws a NumberFormatException:

 Integer.parseInt(s,2);

Could not figure out why. Any idea? The strange thing is if I replace the first bit in the string from the left (1) with a zero bit (0), then it works. The first bit is only for indicating the sign of the number, so I should not make any overflow of 4 bytes in here, right?

=================

UPDATED QUESTION: Thanks to everyone for the answers. I got it now; basically the parseInt() method CANNOT handle the sign bit and it assumes that the sign bit just represents the value as any other bit. So now the question is how to I convert such binary string to a signed INTEGER number? I know some of you suggested I use long, but in my situation I just need it to be an int (long to explain, but you can assume that).

because its binary, not an integer. if it were 1011, you'd expect 11, but you'd get 1,011 — hubson bropa
– hubson bropa, Commented Jun 3, 2015 at 4:02

Nir Alfasi · Accepted Answer · 2015-06-03 04:23:06Z

3

The number that you're trying to parse is bigger than 2^31-1 (max integer), use Long instead:

String s = "11111111110000011101110111011000";
System.out.println(Long.parseLong(s, 2)); // 4,290,895,320

If you want to see what's the maximum value of an Integer you can do:

System.out.println(Integer.MAX_VALUE); // 2,147,483,647

Responding your UPDATE section, if you don't mind loosing information (which happens when you convert long to int) you can do:

String s = "11111111110000011101110111011000";
long l = Long.parseLong(s,2);
int i = (int)l;
System.out.println(i); // -4071976

edited Jun 3, 2015 at 4:23

answered Jun 3, 2015 at 4:01

Nir Alfasi

53.6k11 gold badges94 silver badges138 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Simo Over a year ago

Can you explain why it is bigger than 2^32 while the first bit (from the left) is only used to indicate the sign of the number?

Nir Alfasi Over a year ago

@Simo not sure if I got what you asked - but I added a part to the answer that might explain it.

Simo Over a year ago

Hey alfasin. Got your answer. Thanks. Any idea of how to convert such binary string to an int (not to a long). It has something to do with my encoding for compression, so long numbers will not help with compression.

Nir Alfasi Over a year ago

@Simo you cannot convert it to int without losing information. If you're okay with losing information then you can check the length of s and remove characters until you have a string of length 30 (you should rather avoid strings of length 31 - just to avoid the edge case of hitting 2^31). One way to do it would be as I stated above in the additional example.

Amadan Over a year ago

He's not losing information - he has 32 bits, and wishes to interpret them as a signed int.

|

Amadan · Accepted Answer · 2015-06-03 04:08:01Z

2

You are trying to parse a 32-bit number into an Integer, which can only have 31 bits and a sign. parseInt can't handle the sign bit; you need to give the sign as + or - (or nothing). Your number should be (if I haven't screwed up my conversion) -4071976, which can be parsed if you give it to parseInt as -1111100010001000101000, but not as 11111111110000011101110111011000.

edited Jun 3, 2015 at 4:08

answered Jun 3, 2015 at 4:02

Amadan

200k23 gold badges252 silver badges321 bronze badges

2 Comments

Simo Over a year ago

Got it. Thanks. Now my question is how to convert that binary string to an int number? In my app (has to do with encoding), I need it to be an int, so the first bit needs to represent the sign.

Amadan Over a year ago

(int)Long.parseLong(s, 2) is -4071976 (as a signed int as you want, not as long).

Nir Alfasi · Accepted Answer · 2015-06-03 04:29:43Z

2

As other people already mentioned, this binary string does not fit the 32-bit singed int. However since Java 8 you can parse it as unsigned int, which works perfectly!

String s = "11111111110000011101110111011000";
System.out.println(Integer.parseUnsignedInt(s, 2)); // -4071976

There are other useful unsigned math methods added in Java 8 as well.

edited Jun 3, 2015 at 4:29

Nir Alfasi

53.6k11 gold badges94 silver badges138 bronze badges

answered Jun 3, 2015 at 4:28

Tagir Valeev

101k19 gold badges233 silver badges346 bronze badges

1 Comment

Nir Alfasi Over a year ago

Nice! I wasn't aware of this option!

Estimate · Accepted Answer · 2015-06-03 04:22:55Z

0

You are telling that you are trying to parse binay String

String s = "11111111110000011101110111011000";

But this value is out of Java integer range:

Integer is 32 bit and for Java, the range is -2,147,483,648 to 2,147,483,647. But your input has the value which is out of the range. So you are getting the exception.

But long is having 64 bit and the range is –9,223,372,036,854,775,808 to 9 ,223,372,036,854,775,807. So you can use the below:

String s = "11111111110000011101110111011000";
Long.parseLong(s,2);//4290895320

edited Jun 3, 2015 at 4:22

answered Jun 3, 2015 at 4:04

Estimate

1,4611 gold badge18 silver badges32 bronze badges

Collectives™ on Stack Overflow

NumberFormatException while converting a binary string to a number

4 Answers 4

8 Comments

2 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

8 Comments

2 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related