I replace int s3 by long s3, but it still got a negative .
rossum    

First, that you top-posted.  Please use trim-and-inline posting.

Did you read rossum's message?  Let me quote it:

You know that Java doesn't have unsigned integer types, so why do you ask
what's wrong?  What you call "wrong" is just that Java doesn't have unsigned
integer types.

A couple of things. First of all, you are missing some parentheses:

rt=((i1&0xFFFF)<<16)+(i2&0xffff);

Although it is not necessary in this case, because the leading bit of i1
is zero, in general you need to print it as the low order 32 bits of a
long to avoid negative output:

System.out.println(rt & 0xffffffffL);

May I ask why you are doing all this? Generally, a short[2] is a more
convenient, less fiddly, representation of a pair of short values.

Patricia

 

[Top posting modified.]

You should declare rt as a long, not an int.  As an int rt will only
hold 31 unsigned bits, not 32.  A long will hold up to 63 unsigned
bits.

As Patricia has pointed out you are missing a pair of brackets:

 rt = ((i1 & 0xFFFF) << 16) + (i2 & 0xffff);  
      ^                   ^
I would also be inclined to make at least one of the operands a long,
just to be sure that the addition is done in 64 bits, not 32:

 result = ((i1 & 0xFFFF) << 16) + (long)(i2 & 0xFFFF);  

Look at the bit pattern used to represent both numbers.  Pay especial
attention to the most significant bit of the pattern.  Now look up
what role the most significant bit plays in a 32 bit Java integer.

rossum

Java does not support unsigned.  You used a signed printing routine.
To see it as C does, you would have to use an unsigned printing
routine. The bits are the same for both.  You would have to write your
own unsignedToString method.  You would write a very simple one by
masking off the high 32 bits after a conversion to long, then a
Long.toString.

see http://mindprod.com/jgloss/unsigned.html
for the code.