NumberFormatException to many hex chars

Question

I am receiving an exception when I insert to many hex characters into a string that is being converted to an integer value.

Code:

//variables used in this problem
int INT_MAX = 2147483647;
int INT_MIN = -2147483648;
int currentBase = 16;
string displayValue;

switch(currentBase) {
    case 16:
        if((Integer.valueOf(displayValue + "F", currentBase) >= INT_MIN) && (Integer.valueOf(displayValue + "F", currentBase) <= INT_MAX))
        {
            displayValue += "F";
        }
        break;
}

The error message:

E/AndroidRuntime(6944): Caused by: java.lang.NumberFormatException: Invalid int: "FFFFFFFF"
E/AndroidRuntime(6944):     at java.lang.Integer.invalidInt(Integer.java:138)
E/AndroidRuntime(6944):     at java.lang.Integer.parse(Integer.java:378)
E/AndroidRuntime(6944):     at java.lang.Integer.parseInt(Integer.java:366)
E/AndroidRuntime(6944):     at java.lang.Integer.valueOf(Integer.java:510)
E/AndroidRuntime(6944):     at com.example.calculator.MainActivity.send_f(MainActivity.java:968)

I need this string to have a value no larger than size int (32 bits). Any suggestions?

the number is user input, i am trying to use this value in a calculator that supports doublewords (32 bits). — Taylor, Sep 01 '13 at 03:13
As you seem to already be aware, Java `int` values range from -2147483648 to 2147483647. The hex value 0xFFFFFFFF is 4294967295, which is outside this range. — Greg Hewgill, Sep 01 '13 at 03:15
how could i take 0xFFFFFFFF and use it to represent its negative value (-1)? — Taylor, Sep 01 '13 at 03:18

score 1 · Accepted Answer · edited May 14 '15 at 05:55

1

Greg Hewgill's comment is correct. It looks like you're trying to have the bits be automatically interpreted. Probably the simplest thing to do is something like

 Long.valueOf(displayValue + "F", currentBase).intValue();

to get the bits, and then truncate them to int. Of course, now it's your responsibility to ensure that the argument to Long.valueOf fits in 32 bits.

edited May 14 '15 at 05:55

Stephen Niedzielski

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answered Sep 01 '13 at 03:27

Jeremy

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thank you! i got everything working and now have a much better understanding of how these conversions work! – Taylor Sep 01 '13 at 04:19

score 1 · Answer 2 · answered Sep 01 '13 at 03:29

1

FFFFFFFF is too big, max int value is 0x7FFFFFFF the workaround is

int i = (int)Long.parseLong(hexStr, 16);

answered Sep 01 '13 at 03:29

Evgeniy Dorofeev

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thanks, i understand 0x7FFFFFFFF is the max int value. how would i take 0xFFFFFFFF and get the int value of -1? – Taylor Sep 01 '13 at 03:33
(int)Long.parseLong("FFFFFFFFF", 16) gives -1, try – Evgeniy Dorofeev Sep 01 '13 at 03:59

score 0 · Answer 3 · answered Sep 01 '13 at 03:17

0

try {
    // try your calculation here
} catch (NumberFormatException e) {
    // assign the variable the max/min value instead
    // or whatever behavior you want
}

This code will provide default behaviors for numbers that are out of range.

answered Sep 01 '13 at 03:17

user1549672

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score 0 · Answer 4 · edited May 23 '17 at 11:58

0

It seems to me you are using a wrong method for hex string to int conversion , if my this understand is wrong let me know

I am quoting a duplicate post

    String hex = "ff"

// small values
    int value = Integer.parseInt(hex, 16);  

// big values
    BigInteger value = new BigInteger(hex, 16);

refer for the duplicate post i quoted

API refrence for the method

edited May 23 '17 at 11:58

Community

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answered Sep 01 '13 at 03:21

Srinath Ganesh

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NumberFormatException to many hex chars

4 Answers4