NumberFormatException from hex number [duplicate]

Question

This question already has answers here:

Convert hex string to int (9 answers)

Closed 9 years ago.

In my code I have

int i = Integer.parseInt("f8004896",16);

when I run the program it throws a NumberFormatException

java.lang.NumberFormatException: For input string: "f8004896"

What am I doing wrong? This seems relatively straightforward but its not working.

Answer 1

long i = Long.parseLong("f8004896", 16);
System.out.println(i);
System.out.println(Integer.MAX_VALUE);

Output:

4160768150
2147483647

Answer 2

Decimal value for f8004896 is 4160768150 and it's more than 2^31-1 (upper limit for int type). So you should use long instead: Long.parseLong()