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# making signed int to unsigned int ?

benny rusli
Ranch Hand
Posts: 72
Hello people,

how to make a signed int to unsigned int, i want to show the int number more than 2147483647 for positiv number. We know that int have 4 Byte = 32 bits that means -(2^16)<number<(2^16)-1, i want to make it (2^32)-1 positiv number.

Any help will be appreciated, thanks.

Jeff Albertson
Ranch Hand
Posts: 1780
//for literals:
long n = 2147483648L;
System.out.println(n);

//for negative signed ints you wish to regard as unsigned:
int x = Integer.MIN_VALUE;
//...
long y = 0xffffffffL & x;
System.out.println(y);

Jesper de Jong
Java Cowboy
Saloon Keeper
Posts: 15568
46
Java does not have unsigned integers. So as Jeff shows, you can use a larger type (a long, which is 64 bits) to hold values that don't fit into int (which is 32 bits).

benny rusli
Ranch Hand
Posts: 72
Hello people,

can someone explain me, what is the meaning of this row from the code below : "String s = Integer.toHexString( digest[i] & 0xFF );" , what is digest[i] & 0xFF. When to use byte and when to use int ? whether the both is same ? if i print out the byte = 20; is the same if i print out int = 20;. Many thanks for responding my question.

Layne Lund
Ranch Hand
Posts: 3061
Originally posted by arifin rusli:
Hello people,

can someone explain me, what is the meaning of this row from the code below : "String s = Integer.toHexString( digest[i] & 0xFF );" , what is digest[i] & 0xFF. When to use byte and when to use int ? whether the both is same ? if i print out the byte = 20; is the same if i print out int = 20;. Many thanks for responding my question.

First of all, digest[i] gets the ith element from an array named digest. 0xFF is a int constant. The "0x" at the beginning means that it is represented in hexadecimal (hex), which is popular for representing binary numbers. The hex value of FF is 255. If you need more details on how this works, you should google for information on number systems.

Finally, the & operator performs a bit-wise AND operation. That is, it takes the corresponding bits in each of the operands. If both bits are 1, then the resulting bit is 1. Otherwise, the resulting bit is 0. Perhaps an example will make this more clear: