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Answer 1

+4 A:

First thing is you'll need a 64 bit value to store it in. Assuming your callId values are (32 bit) ints, you'd need to do something like this.

int callId1, callId2;
...
long globalCallId = ((long)callId1 << 32) + callId2;

<< is the bit shift operator - shifting 32 bits is the equivalent of multiplying by 2^32.

David M 2010-06-15 11:53:19

Thanks a bunch, really helpfull.

Jonas B 2010-06-15 11:59:30

Answer 2

+1 A:

It's easiest to shift callid1 by 32 bits.

long globalCallId = ((long)callid1 << 32) + callid2;

Stefan Steinegger 2010-06-15 11:53:59

Please note that this will only work if both callid1 and callid2 are longs. Explicit casting to long is required if they are ints. You have to be careful of overflows while using shift operations.

apoorv020 2010-06-15 11:58:29

@apoorv020: It's about the calculation here, not about the types. Who said that they aren't long? If you prefer, I'll cast it.

Stefan Steinegger 2010-06-15 12:15:55

Answer 3

A:

So global callid = callid1 * 2³² + callid2. You can use:

long globalCallID = (callid1 << 32) + callid2

This uses the fact that a << b == a multiplied by 2 to the power of b.

IVlad 2010-06-15 11:54:20

Please note that this will only work if both callid1 and callid2 are longs. Explicit casting to long is required if they are ints. You have to be careful of overflows while using shift operations.

apoorv020 2010-06-15 11:59:04

@apoorv020: Please stop downvoting all the answers which miss the cast. You don't know if it is actually needed.

Stefan Steinegger 2010-06-15 12:18:52

Maybe I overdid it a bit. I apologize.

apoorv020 2010-06-15 12:28:20

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c# calculation help (TExtractor callid)

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