try to give the logic

[kad]

Give a very good method to count the number of ones in a 32 bit number. (caution: looping through testing each bit is not a solution).

 

[fenugrec]

What ? Is that an order?

I hope you don't expect anyone to "give" you a ready-made solution...

google for "parity"

 

[electronrancher]

not an order, just verbatim from the homework assignment - lol!

 

[tronix]

not an order, just verbatim from the homework assignment - lol!

Sure looks like a straight lift but cann't resist frm replying

let x be d no. not equal to zero

ans=1
while(x>2)
{
if( x%2!=0) {ans++; x--;}
x=x/2;
}

ans will give the no. of ones

TRY TO UNDERSTAND THE LOGIC AS THATS WHAT MATTERS

 

[polymath]

Hi all

On average there will be 16 - the most probable prediction every time.

Polymath

 

[tronix]

Hi all

On average there will be 16 - the most probable prediction every time.

Polymath

Why should ?

 

[polymath]

Hi all

On average there will be 16 - the most probable prediction every time.

Polymath

Why should ?

Why not? You have not defined your 32 bits as having any structure other than random.

Read about Statistics & Probability.
Noise - Long Comm. Transmissions and many data structures have equal 1s & 0s by chance or design.

yes .. no .. yes .. no ... Polymath

 

[tronix]

Well polymath I think here we r talking abt deterministic coding and not the random ones.

 

[cfant]

You can use lookup table which contains number of ones in one byte. Since the word is 4 bytes (assuming normal cpu), just add results from all four bytes. Or you can use some of these algorithms:
https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetTable