Convert int of unknown number of bits to 32 bits int

Hi all,

need some help, trying to convert a integer of unknown number of bits to a 32 bit.
For example: 2196550012

Can you some how drop the least significant bits after the 32nd bit?
So you get a new integer with only 32 bits, this I assume will get me a new value for the integer, but I hope it will be approximately the same as before.

Can't find any threads on on this subject, you are welcome to direct me towards one =)

Thank you!

Do you need to covert a String to integer?

By the way "2196550012" can be represented with log2(2196550012) = 31.03.... => 32 bits.

You have 2 options:

1) Use library routines. You probably need Integer.parseInteger(String, int).

2) Write conversion code yourself (This is particularly true if this is a learning exercise and you are just learning to program. Tutors usually ask to do things this way.):

say I have a number: "1234" it is: 4 * 10^0 + 3 * 10^1 + 2 * 10^2 + 1 * 10^3. ^ stands for power.

So write a loop from 0 and up and sum products of decimal digits and powers of 10. (If the number is decimal.)

Nope it is not a string!
Is looking at different Ip lookup algoritms in a project, and for testing these I shall use real data from existing routing tables, and the number given as example is a destination address converted to an integer see ex. http://www.aboutmyip.com/AboutMyXApp...30.236.177.124

And have been provided with a file with already converted addresses, so would like to convert these numbers to 32 bits int.

[QUOTE=andrey_zh; By the way "2196550012" can be represented with log2(2196550012) = 31.03.... => 32 bits.
[/QUOTE]

Okey, but why does the Eclipse complain about that it out of range? Because it only use up to 2^31-1 bits? What I have read Java does not support unsgined!

2196550012_10 = 82ECB17C_16

82_16 = 130
EC_16 = 236
B1_16 = 177
7C_16 = 124

Seems you have to use masks if you need to split an integer to four octets.

a.b.c.d

Code:

---

a = (ip >> 24) & 0xFF;
b = (ip >> 16) & 0xFF;
c = (ip >>  8) & 0xFF;
d = (ip      ) & 0xFF;

---

And yes some IP-s will be stored as negative integers.

Tip: read as long, then convert to int:

Code:

---

long a = 2196550012L;
System.out.println( (int)a );

---

Gives the result: -2098417284_10 = 82ECB17C_16.

thank you, will try that =) at once.

Coz 2196550012 is out of range of int type! do like that (int)(2196550012L). If you have it in a file either read as long or as a String, then convert.

Quote:

---

-2098417284_10 = 82ECB17C_16.

---

This is true for 32 bit representation, not mathematically of course.