What is the purpose of "ts & 0xffff0000"?

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I'm working on a real time protocol that adding the timestamp for each transmitted packet and I don't understand what the lines of code mean. Thanks for help.

// ts for timestamp
unsigned int ts;


if(ts & 0xffff0000){
 // do something
}
c
network-programming
operators
asked on Stack Overflow Jul 23, 2014 by user2958240 • edited Jul 23, 2014 by user2864740

4 Answers

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Given the fact they're using binary-and (&), the intent seems to be to check if any of the 16 high bits are set.

Binary-and examines the bits at each position in both numbers, and if they're both 1, then the result as a 1 bit in that same position. Otherwise the result has a zero in that position

0b 001001001001001001001001001001  (first number, usually a variable)
0b 010101010101010101010101010101  (second number, usually a "mask")
=================================
0b 000001000001000001000001000001  (result)

If this is used as the condition of an if-block, such as if (x & mask), then the if-block is entered if x has any of the same bits as mask set. For 0xFFFF0000, the block will be entered if any of the high 16 bits are set.

That is effectively the same as if (ts > 65535) (if int is 32bit or less), but apparently the intent is to deal with bits, rather than the actual value.

answered on Stack Overflow Jul 23, 2014 by Mooing Duck • edited Jul 23, 2014 by Mooing Duck
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0xffff0000 serves as a bit mask here.

ts & 0xffff0000 satisfies as a condition when some bit in the first 16 bits of ts is 1. Put another way, when ts >= 2^16.

answered on Stack Overflow Jul 23, 2014 by Peng Zhang
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This IF loop checks if any of upper 16 bits of ts is high. If yes, then loop is executed.

The IF loop is executed only if ts >= 0x00010000.

answered on Stack Overflow Jul 24, 2014 by rohit • edited Jul 24, 2014 by rohit
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An intuitive way to understand this.

    **** **** **** **** //the first 16bits of ts
&   1111 1111 1111 1111 //the first 16bits of 0xffff 0000

If one of the first 16bits of ts is set, then the result above won't be zero.
If they are all 0, the result above will be 0000 0000 0000 0000

For the last 16bits of ts, no matter what happens to these bits, the result of Binary-and will be 0.

    **** **** **** ****
&   0000 0000 0000 0000
   =0000 0000 0000 0000

So if the first 16 bits of ts have one 1 bit ==> ts&0xffff0000 > 0 (which means ts>=0b 10000 0000 0000 0000(i.e., 2^16)), el se ts&0xffff0000 == 0.

Always, we also use this ts&1 to test whether ts is an odd number.

answered on Stack Overflow Dec 8, 2020 by youpen • edited Dec 8, 2020 by youpen

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