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Bit shifting?

 
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answer 2,4
here how can the answer be 4...
my doubt is here the long variable after moving ..how can it be assigned it to int ...without explicit casting ....
please clarify

thanks
sri
 
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The long is *not* being assigned to anything. It is merely being used to to state how many bits to shift. And even in this regard, only the lower 5 bits of the long will be used.

No assignment of the long. Hence, no casting necessary.

Henry
 
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Originally posted by Henry Wong:
And even in this regard, only the lower 5 bits of the long will be used.



What's meant with that sentence?

because i tought the number of bits that will be shifted is 3 (99 % 32 (= max bits of an int)), so have no idea what's meant with lower 5 bits of long will be used
 
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because i tought the number of bits that will be shifted is 3 (99 % 32 (= max bits of an int)), so have no idea what's meant with lower 5 bits of long will be used



99 & 31 = 3

99 in binary:
00000000 00000000 00000000 01100011

31 in binary:
00000000 00000000 00000000 00011111

When they are ANDed together, the result is (in binary):
00000000 00000000 00000000 00000011

Or, 3. This is how everything but the five lowest bits are masked off. It's mathematically equivalent to %32, but the JLS specifically describes the behavior as using the five lowest bits for int values (six lowest for long).

Hope this helps...
 
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And for those of you studying for the Tiger exam, remember this topic doesn't apply to you, it's not on the Tiger exam!
 
Roel De Nijs
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Originally posted by Steve Morrow:
Hope this helps...



it did help.
thanks!
 
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