Convert Java String hexadecimal number to int hexadecimal number

Question

I need to figure out, how to set the hexa number 0xffff, which is used for bitwise AND in code below dynamically.

int offsetData = address & 0xffff;

For example I need something like this:

int offsetData = address & myHexaValue;

I have a decimal int value, which I need to put in hexadecimal format instead of myHexaValue.

This is my state now.

int decimalShift = 16*8;    
String myHexaValueString = Integer.toHexString(decimalShift);

Now.. can I convert my String hexa number to int hexa number so I could put result into myHexaValue?

Maybe I dont understand, but how it solve my problem? I need to set this hexa value dynamically. Method is obtaining only decimal value. — Daniel Perník
– Daniel Perník, Commented Jan 17, 2016 at 20:30

someone_somewhere · Accepted Answer · 2016-01-17 21:07:25Z

2

You don't need to convert it. There is no such thing as a hex number. A number written in hex is a number written in a base 16 notation. We normally use base 10 notation so we have digits from 0 to 9 but in hex notation you have digits from 0 to F. So just do a and operation :

int offsetData = address & decimalShift;

EDIT: To get only the lower 16 bit use a mask. 0xffff and perform a & operation the upper 16 bits will be set to 0 and the lower 16 will remain unchanged. so:

int offsetData = address & decimalShift & 0xffff;

EDIT2: To get a mask for any number(0,...,31) of bits you can use this function:

private static int getMask(int numOfOnes) {
        int i = 0x80000000;
        i = i>>(32-numOfOnes-1);
        i = ~i;
        return i;
    }

Pauls explanation: "bitshifting with >> fills the most signifcant bits with the value of the MSB before shifting, so in this code all bits except for the numOfOnes lowest bits are filled with 1s and afterwards the number is inverted."

edited Jan 17, 2016 at 21:07

answered Jan 17, 2016 at 20:29

someone_somewhere

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5 Comments

Daniel Perník Over a year ago

My goal is to get "some number" of last bits of (32 bit) integer. If it will be 16 bites, solution will be int val = x & 0xffff; ... I need to figure out, how to get n numbers.

Daniel Perník Over a year ago

Please, can you show me what will change in code, if I will need for example 8 and 5 lower bits? Thank you. I still don't know, what purpose have here & decimalShift.

user4668606 Over a year ago

note concerning EDIT2: bitshifting with >> fills the most signifcant bits with the value of the MSB before shifting, so in this code all bits except for the numOfOnes lowest bits are filled with 1s and afterwards the number is inverted.

someone_somewhere Over a year ago

I added your (@Paul) explanation as part of the answer so it is more noticeable.

user4668606 Over a year ago

@someone_somewhere thanks, hoped you'd do that, just didn't want to edit the answer myself, since it's a pretty nice answer after all. But since OP obviously isn't that adept with the number-system and bit-operations I thought this should be part of the answer.

Bohemian · Accepted Answer · 2016-01-17 21:45:10Z

1

Use Integer.parseInt(), specifying a radix of 16:

int number = Integer.parseInt(myHexaNumber, 16);

answered Jan 17, 2016 at 21:45

Bohemian♦

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Comments

user4668606 · Accepted Answer · 2016-01-17 21:01:23Z

0

Numbers aren't "hexadecimal" or "decimal". This is just an imprecise spelling that is used because of laziness. Numbers are represented decimal or hexadecimal would be precise. There's no need to convert an int from decimal to hexadecimal. In fact they're stored in binary-format, no matter what notation you use in the code. The only point where notation is relevant is for user-interactions or during coding, like in- and outputting numbers, or specifying constants in the code.

answered Jan 17, 2016 at 21:01

user4668606

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