Parsing a Hexadecimal String to an Integer throws a NumberFormatException?

Question

So, In Java, you know how you can declare integers like this:

int hex = 0x00ff00;

I thought that you should be able to reverse that process. I have this code:

Integer.valueOf(primary.getFullHex());

where primary is an object of a custom Color class. It's constructor takes an Integer for opacity (0-99) and a hex String (e.g. 00ff00).

This is the getFullHex method:

public String getFullHex() {
    return ("0x" + hex);
}

When I call this method it gives my this NumberFormatException:

java.lang.NumberFormatException: For input string: "0xff0000"

I can't understand what's going on. Can someone please explain?

Oliv · Accepted Answer · 2016-06-13 07:43:55Z

98

Will this help?

Integer.parseInt("00ff00", 16)

16 means that you should interpret the string as 16-based (hexadecimal). By using 2 you can parse binary number, 8 stands for octal. 10 is default and parses decimal numbers.

In your case Integer.parseInt(primary.getFullHex(), 16) won't work due to 0x prefix prepended by getFullHex() - get rid of and you'll be fine.

edited Jun 13, 2016 at 7:43

Oliv

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answered Jul 7, 2012 at 19:20

Tomasz Nurkiewicz

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4 Comments

mattbdean Over a year ago

Thank you! I didn't know that Integer.parseInt(..) could take another parameter! Thanks for clearing that up for me!

user3402040 Over a year ago

not work java.lang.NumberFormatException: Invalid int: "0x920B540C", color2 = Integer.parseInt(color_2,16); (with argbA)

Remy Lebeau Over a year ago

Even if you remove the 0x prefix from the input string "0x920B540C" and parse it in base 16, hex 920B540C is decimal 2450215948, which is more than the max value that a 32bit signed int can hold (dec 2147483647, hex 0x7FFFFFFF - see Integer.MAX_VALUE). You would have to use a 64bit signed long instead (Java doesn't have unsigned integer types).

Line Over a year ago

I came here searching for an elegant solution for getting rid of 0x. So -1 for not mentionig Integer.decode().

Konstantin Labun · Accepted Answer · 2016-08-16 13:15:10Z

47

Try to use decode method:

Integer.decode("0x00ff00");

DecodableString:

Signopt DecimalNumeral
Signopt 0x HexDigits
Signopt 0X HexDigits
Signopt # HexDigits
Signopt 0 OctalDigits

You can read about it here https://docs.oracle.com/javase/7/docs/api/java/lang/Integer.html#decode(java.lang.String)

answered Aug 16, 2016 at 13:15

Konstantin Labun

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2 Comments

Pablo Santa Cruz Over a year ago

Best answer IMHO.

Ben Kushigian Over a year ago

Just adding that this doesn't handle binary (0b...) prefixes

nos · Accepted Answer · 2012-07-07 19:24:58Z

39

Integer.valueOf(string) assumes a decimal representation. You have to specify that the number is in hex format, e.g.

int value = Integer.valueOf("00ff0000", 16);

Note that Integer.valueOf(string,16); does not accept a prefix of 0x. If your string contains the 0x prefix, you can use Integer.decode("0x00ff0000");

answered Jul 7, 2012 at 19:24

nos

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6 Comments

Ashish Sahu Over a year ago

Hey this is working only for 00ff0000 not for 80ff0000 . I have string like String hex="803BB9FF"; and i want to covert this into int color=0x803BB9FF please help

nos Over a year ago

@AshishSahu It's impossible to help when you don't describe what happens, and what you expect to happen. 0x803BB9FF is -2143569409 (since int in Java is signed). So what's "not working" about -2143569409 ?

Ashish Sahu Over a year ago

I have 8char color string like "803BB9FF" now i want to convert it in int value like 0x803BB9FF (int) i tryied with "003bb9ff " then it's working but not working with 80 prefix

Ashish Sahu Over a year ago

But thanks for reply my vote+. now i got another solution : int color=Color.parseColor("#"+"803bb9ff");

Krease Over a year ago

Integer.decode was exactly what I needed! - it's basically parseInt but handles more variety of input formats

|

Roger Lindsjö · Accepted Answer · 2012-07-07 19:24:10Z

3

The parseInt method only accepts the number part, not any kind of "base" indicator such as "0x" for hexadecimal or "0" for octal. Use it like this

int decimal = Integer.parseInt("1234", 10);
int octal = Integer.parseInt("1234", 8);
int hex = Integer.parseInt("1234", 16);

answered Jul 7, 2012 at 19:24

Roger Lindsjö

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Comments

Robert Columbia · Accepted Answer · 2019-08-07 20:49:00Z

3

Would this work?

Long.parseUnsignedLong(value, 16);

edited Aug 7, 2019 at 20:49

Robert Columbia

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answered Aug 7, 2019 at 20:38

Matt

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1 Comment

indigo Over a year ago

This worked for me. I was working with "800b6b0b": 2,148,231,947 Just over the 32 bit limit

GETah · Accepted Answer · 2012-07-07 19:23:45Z

0

This should do it:

String hex = "FA"; 
int intValue = Integer.parseInt(hex, 16);

And if you want to generate hex representation of an integer, use

String hex = Integer.toHexString(12);

answered Jul 7, 2012 at 19:23

GETah

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Comments

Manuel_B · Accepted Answer · 2025-11-21 15:08:25Z

0

In newer Java Versions (since 17) there is HexFormat:

     byte byteVal = (byte)HexFormat.of().fromHexDigits("7f");
     assert(byteStr.equals("7f"));
     assert(b == byteVal);

     HexFormat commaFormat = HexFormat.ofDelimiter(", ").withPrefix("#");
     byte[] bytes = {0, 1, 2, 3, 124, 125, 126, 127};
     String str = commaFormat.formatHex(bytes);

     byte[] parsed = commaFormat.parseHex(str);
     assert(Arrays.equals(bytes, parsed));

     // The formatted string is: "#00, #01, #02, #03, #7c, #7d, #7e, #7f"

https://docs.oracle.com/en/java/javase/21/docs/api/java.base/java/util/HexFormat.html

answered Nov 21 at 15:08

Manuel_B

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