Java integer operations

Question

My int values:

    int a= 0x02; int b= 1; int c= 2; int d = 0; int e = 0; int f = 0; int g= 1;

How can I concatenate these integers and get a single int value? and the result int value should have first 8 bits for "a" and 4 bits each for the rest??

Something like : 2120001

"Something like : 2120001" Do you expect that result as a decimal number? That contradicts your first sentence where you talk about bits. So what result exactly do you expect? — Jesper
– Jesper, Commented Dec 5, 2012 at 9:25
WHat are you trying to do because I suspect this won't work the way you think it does. For example 0x02 is just 2 and there is no way to determine the number was defined in hex. — Peter Lawrey
– Peter Lawrey, Commented Dec 5, 2012 at 9:26
You don't really "concatenate" ints. You add ints, and concatenate strings. You could convert each of your ints to a string, concatenate them, and convert back to an int, but that result is not even guaranteed to be an int. — The111
– The111, Commented Dec 5, 2012 at 9:26

pbhd · Accepted Answer · 2012-12-05 09:36:20Z

3

int a= 0x02; int b= 1; int c= 2; int d = 0; int e = 0; int f = 0; int g= 1;
int res=((a&0xff)<<24)|((b&0xf)<<20)|((c&0xf)<<16)
       |((d&0xf)<<12)|((e&0xf)<<8)|((f&0xf)<<4)|((g&0xf));
System.out.println(Integer.toHexString(res));

Yields: 2120001

answered Dec 5, 2012 at 9:36

pbhd

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1 Comment

user207421 Over a year ago

It doesn't yield 212001. It yields 0x212001. Not the same thing.

Ajinkya · Accepted Answer · 2012-12-05 09:27:27Z

3

String yourString = "" + a + b + c + d + e + f + g;
int finalInt = Integer.parseInt(yourString);

edited Dec 5, 2012 at 9:27

Ajinkya

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answered Dec 5, 2012 at 9:26

Michael 'Maik' Ardan

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6 Comments

Joop Eggen Over a year ago

Integer.parseInt(yourString, 4);

The111 Over a year ago

This will not meet the specs of the question, as far as the first 8 bits, the next 4 bits, etc. For example, the first 8 bits of this resulting int will be 00000000.

Michael 'Maik' Ardan Over a year ago

0x02 = 2. As per the problem, 0x02,1,2,0,0,0,1 = 212001 which is the poster's desired result.

Michael 'Maik' Ardan Over a year ago

int a = 0x02; int b = 1; int b = 2; int d = 0; int e = 0; int f = 0; int g = 1; String z = "" + a + b + c + d + e + f + g; int j = Integer.parseInt(z); System.out.println(j);

TheBlastOne Over a year ago

0x02,1,2,0,0,0,1 = 2120001, not 212001, @MichaelArdan

|

Joop Eggen · Accepted Answer · 2012-12-05 09:44:55Z

1

Forget

With thanks to @MichaelArdan

String yourString = "" + b + c + d + e + f + g;
int finalInt = Integer.parseInt(yourString, 4) | (a << (6*2));

It interpretes bcdefg as base 4 number, and shifts a to the end.

(Misread the question)

int n = a;
int[] v = new int[] { b, c, d, e, f, g };
int p = 8;
for (int k : v) {
     n |= (k & 0xF) << p;
     p += 4;
}
System.out.println(n);

edited Dec 5, 2012 at 9:44

answered Dec 5, 2012 at 9:32

Joop Eggen

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2 Comments

Michael 'Maik' Ardan Over a year ago

:D int a = 0x02 would still be considered as 2.

TheBlastOne Over a year ago

Here, I also don´t like the using-string-concatenation-operator-to-avoid-doing-the-arithmetic-right nature of the answer.

j0k · Accepted Answer · 2012-12-05 09:30:02Z

0

How about

String s = "" + a + b + c + d + e + f + g;
int concat = Integer.parseInt(s);

and read up on the Java type conversion and promotions.

edited Dec 5, 2012 at 9:30

j0k

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answered Dec 5, 2012 at 9:24

Anders R. Bystrup

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3 Comments

vishal_aim Over a year ago

what about "result int value should have first 8 bits for the "a" and the 4 bits each for the rest??"

Ajinkya Over a year ago

Should convert back "s" it to int.

TheBlastOne Over a year ago

I don´t like this since you are doing a lot of StringBuilder calls between-the-lines just to avoid some highly efficient bitwise operations like shifts and ors (or adds). A division can be implemented as a series of string manipulations, too, but that would be far from efficient, and far away from the true nature of the operation.

Bhavik Ambani · Accepted Answer · 2012-12-05 09:31:08Z

0

You can achieve this by the following way

public static void main(String[] args) throws UnsupportedEncodingException {
        int a = 5;
        int b = 3;
        int bLength = Integer.toBinaryString(b).length();
        System.out.println("a = "  + Integer.toBinaryString(a));
        System.out.println("b = "  + Integer.toBinaryString(b));
        int c = (a << bLength) | b;

        //System.out.println(c);
        System.out.println("c = "  + Integer.toBinaryString(c));


    }

Output

a = 101
b = 11
c = 10111

answered Dec 5, 2012 at 9:31

Bhavik Ambani

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Comments

TheBlastOne · Accepted Answer · 2012-12-05 09:33:44Z

0

I think what you want is something like this:

public static void main(String[] args)
    {
        int a=2;
        int b=1;
        int c=2;
        int d=0;
        int e=0;
        int f=0;
        int g=1;

        System.out.println(g+(f << 4)+(e << 8)+(d << 12)+(c << 16)+(b << 20)+(a << 28));
    }

Result is 538050561, which is 20120001 in hex.

See also http://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html for bitwise shift operators.

answered Dec 5, 2012 at 9:33

TheBlastOne

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Comments

user207421 · Accepted Answer · 2012-12-05 23:19:48Z

0

To yield 212001 you need

((((((a*10)+b)*10+c)*10+d)*10+e)*10+f)*10+g

which shows you what's going on, or more simply

a*1000000+b*100000+c*10000+d*1000+e*100+f*10+g

However that contradicts your statement about bit width.

I conclude that what you really want is 0x212001, which is yielded by

(((((((((((a << 4)+b) << 4)+c) << 4)+d) << 4)+e) << 4)+f) << 4)+g

or more simply

(a << 24)|(b << 20)|(c << 16)|(d << 12)|(e << 8)|(f << 8)|g

answered Dec 5, 2012 at 23:19

user207421

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