C/C++: Get Help on Unsigned Int x &= x-1;

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SUMMARY

The operation x &= x-1 in C/C++ effectively clears the least significant set bit (1) in the unsigned integer x. This works because the expression performs a bitwise AND between x and x-1, which has all bits lower than the least significant set bit of x set to 1. This method is advantageous for counting set bits in an unsigned integer as it reduces the number of iterations compared to looping through all bit positions, making it more efficient.

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andytran
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Hi,

After hrs of googling still couldn't find the answer to my question. I'm hoping i get better luck positng it here. Any C/C++ out there please help!

1)
a) Given an unsigned int x, what does x &= x-1; do and why does it work?

I know what x &= x-1 does but to explain why does it work, i have no clue.

b) How can the statement in a) be used in a C++ program to count set bits (ones) in a given unsigned int x? What is the advantage over looping through all bit positions?


thanks in advance.
 
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For starters, tell us what you already know. What does the & operator do in C?

- Warren
 
Well for 2b), think about how you would loop through the bit positions. What are you going to change for each iteration of the loop? Also think about the different ways that you could do this counting problem, and whether this &= method would have advantages over other ways... BTW, I think there is at least one faster way to do the counting...can you think of it too?
 
chroot said:
For starters, tell us what you already know. What does the & operator do in C?

- Warren

& is the bitwise and operator, x &= x-1; is the same as x = x & (x-1);

if an unsigned int x = 5; then

5 = 101 in binary and (x-1) = 4 = 100

so after evaluating, x = 100 in binary or x = 4 in decimal.
 

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