Please explain the code on how does bit wise operator works..also Endianness

Question

Hi All I am trying to understand shift operator which is used in my project. What I came to know from google is that Bit shifting allows for compact storage of similar data as a single integral value. But how does that work. and what is the advantage. Can you explain me the part below program whats exactly happening in below lines. The code is taken from http://www.codeproject.com/KB/graphics/dicomImageViewer.aspx (Tnx to Amarnath S) int vr = (b0 This programe reads a binary file in medical terms a dicom file which is basically a binary file. Getting 4 bytes from a dycom file like this: byte b0 = GetByte(); // GetByte() is a private method that returns a single byte. byte b1 = GetByte(); byte b2 = GetByte(); byte b3 = GetByte(); Then doing this: int vr = (b0 Then a switch statement: switch(vr) { // Again didn get whats happening in case UT and QQ // where UT = 0x5554 is a constant declared above in the file case UT: // Explicit VR with 32-bit length if other two bytes are zero if ((b2 == 0) || (b3 == 0)) return GetInt(); // Implicit VR with 32-bit length vr = IMPLICIT_VR; if (littleEndian) return ((b3 else return ((b0 // break; // Not necessary case QQ: // where QQ = 0x3F3F is a constant declared above in the file // Explicit vr with 16-bit length if (littleEndian) return ((b3 else return ((b2 } Can someone please give me some idea on this. What exactly might be happening in the lines having shift operators. am little confused with bits and bytes.. hoping to get some reply. Thanks a lot.

Suraj Rai · Answer

Thanks a lot Karthick and Vulpes Regards

Vulpes · Answer

An int requires 4 bytes of storage and 'little endian' means that those bytes are ordered with the least significant one first. In the case of 'big endian' the most significant byte is stored first. The x-86 processor family (32 or 64 bit) use little endian. If you have 4 bytes: b0 b1 b2 and b3 and you want to combine them into a little endian integer then remembering that b3 is the most significant you need to multiply: b3 by 2 ^ 24 b2 by 2 ^ 16 b1 by 2 ^ 8 b0 by 2 ^ 0 (which is 1 and therefore b0 is unchanged) where 'a ^ b' denotes raising 'a' to the power 'b'. You then add the results together. As Karthik has already explained you can multiply an integer by 2 ^ n by left shifting its bits by 'n' places. So b3 * (2 ^ 24) can be calculated as b3 If you wanted to combine them into a big endian integer then b0 would now be the most significant byte and would therefore need to be left shifted 24 positions and so on for the other bytes.

Karthik Agarwal · Answer

No no you mistook my words it's not that way. With the example you stated everything is correct except you are multiplying with the same number 5 do the multiplication with 2 instead for 4 number of times. i.e int vr = 5; int result = vr int result = vr * 2 * 2 *2 *2; i.e 5 * 2 * 2 * 2 * 2;//80 is the result. same way do it with 65535 * 2 * 2 * 2 ........ so on 8 times which gives 16776960. in the same way

Suraj Rai · Answer

Thanks a lot for the reply. but i am not clear of few things. what do you mean by 8 times multiply..do you mean..say i have int vr = 5; int result = vr is it mean : 5 * 5 * 5 * 5 ?? because when i tried to multiply ur example e.i 65535 it does'n give 16776960. sorry for asking you again i guess i am getting it wrong. Please explain a bit more as I am new to computing lang. And can you tell me about little endian ? I mean why if little endian they are doing leftshift byte wise 24 168 .. if (littleEndian) return ((b3 Thanks. Regards

Karthik Agarwal · Answer

for example if you consider binary data 11111111 which is nothing but 0xFFFF in hexadecimal value. but either you take binary decimal or hexadecimal the operation is one and the same it multiplies the number with 2 that's it. int vr = 11111111 or 0xFFFF or 65535; int result = vr the variable result will be holding a value 16776960 which is like multiplying the number (i.e 11111111 or 0xFFFF or 65535) eight number of times. That is what is done in the above code. In the similar way there is right shift as well >> which does the inverse operation of left shift i.e it does division those many number of times based on the index beside that. ex: int vr = 16776960; int result = vr >> 16776960;// result will have 65535 For any further doubts feel free to post back.