将十六进制字符串解析为int值时发生意外的NumberFormatException

Question

I want to parse an String containing 8 hex-digits (4bytes) but i got an NumberFormatException. What is wrong here?

assertThat(Integer.parseInt("FFFF4C6A",16),is(0xFFFF4C6A));

Answer 1

Your number represents a number greater than that assignable to an int. Try:

Long.parseLong("FFFF4C6A", 16);

which gives 4294921322.

From the doc :

An exception of type NumberFormatException is thrown if any of the following situations occurs:

• The first argument is null or is a string of length zero.
• The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX .
• Any character of the string is not a digit of the specified radix, …
• The value represented by the string is not a value of type int.

and it's the 4th case that you're hitting.

Answer 2

That is because the Integer.parseInt("FFFF4C6A",16) provided exceeds Integer.MAX_VALUE which is defined as public static final int MAX_VALUE = 0x7fffffff;

Now, as per the Javadoc for parseInt(...), you would hit a NumberFormatException in either of the following cases:

An exception of type NumberFormatException is thrown if any of the following situations occurs:

• The first argument is null or is a string of length zero.
• The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
• Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\-') or plus sign '+' ('\+') provided that the string is longer than length 1.
• The value represented by the string is not a value of type int.

In your case, since the String value supplied exceeds Integer.MAX_VALUE, you're satisfying the 4th clause for NumberFormatException

Possible Solution: In order to parse this, use Long.parseLong(...) where the MAX_VALUE is defined as `public static final long MAX_VALUE = 0x7fffffffffffffffL

Answer 3

You've exceeded the range of an integer.

Integer.MAX_VALUE = 2147483647
0xFFFF4C6A = 4294921322

Parsing it as a Long works:

Long.parseLong("FFFF4C6A",16)

Answer 4

如果仅要将十六进制字符串表示为整数（因为它是32位），则需要使用BigInteger ：

new BigInteger("FFFF4C6A", 16).intValue()

Answer 5

I don't know the assertThat() method, but your hexadecimal number "FFFF4C6A" is to big for an integer.

For example, if you write :

int number = Integer.parseInt("FFFF4C6A",16)

you'll get the same error. A correct way to write the code would be :

double number = Integer.parseInt("FFFF4C6A",16)