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what is this mean ">>"?

Posted on 2006-11-14
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Last Modified: 2011-09-20
i have this code from my old colleague, can anyone tell me what does this means?

 
abyLic[4] = (byte) (iLicenseID);
        abyLic[5] = (byte) (iLicenseID >> 8);
        abyLic[6] = (byte) (iLicenseID >> 16);
        abyLic[7] = (byte) (iLicenseID >> 24);
       
        abyLic[8] = (byte) (iSMSCounter);
        abyLic[9] = (byte) (iSMSCounter >> 8);


the abyLic[] is an array and the iLicenseID is int.

what doest that mean by iLicenseId >> 8 ??
is that he is trying to set the value to 8 in abyLic[5]?
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Question by:taiping
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by:CEHJ
ID: 17945013
Shift right operator

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by:Ajay-Singh
ID: 17945014
bitwise shift
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by:CEHJ
ID: 17945018
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girionis earned 500 total points
ID: 17946486
Just to fill up with what the other experts suggested

>what doest that mean by iLicenseId >> 8 ??

The ">>" is a binary operator and it means "right shift the value in the "iLicensedId" eight times". The value of "iLicencedId" is an integer. When you right shift a value you make it smaller. For example if the "iLicencedId" holds the value of 512, which in *binary* it is 000100000000, and you right shift 8 times, it becomes 000000000010 (=2 in decimal). As you see you can see the digit 1 is moved 8 positions to the right. This is what the >> operator is doing, it moves the digits to the right.

> is that he is trying to set the value to 8 in abyLic[5]?

No, he is setting the *converted* value (the one that was shifted to the right 8 times), i.e. the result of the right-shift operation, to the 6th position os the "abyLic" array
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by:girionis
ID: 17946495
Sorry the 512 binary should be 001000000000.
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by:CEHJ
ID: 17953872
taiping - why did you ignore previous correct answers?
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