hex to int number format exception in java

Question

I am getting a number format exception when trying to do it ``` int temp = Integer.parseInt("C050005C",16); ``` if I reduce one of the digits in the hex number it converts but not otherwise. why and how to solve this problem?

Accepted Answer

This would cause an integer overflow, as integers are always signed in Java. From the documentation of that method (emphasis mine):

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 '-' ('\u002D') provided that the string is longer than length 1.

- The value represented by the string is not a value of type int.

It would fit into an unsigned integer, though. As of Java 8 there's Integer.parseUnsignedInt (thanks, Andreas):

```
int temp = Integer.parseIntUnsigned("C050005C",16);
```

On earlier Java versions your best bet here might to use a `long` and then just put the lower 4 bytes of that `long` into an `int`:

```
long x = Long.parseLong("C050005C", 16);
int y = (int) (x & 0xffffffff);
```

Maybe you can even drop the bitwise "and" here, but I can't test right now. But that could shorten it to

```
int y = (int) Long.parseLong("C050005C", 16);
```

hex to int number format exception in java

Solution

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