c++ - Counting number of bits: How does this line work ? n=n&(n-1); -

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i need explanation how specific line works.
know function counts number of 1's bits, how line clears rightmost 1 bit?

int f(int n) {     int c;     (c = 0; n != 0; ++c)          n = n & (n - 1);     return c; }

can explain me briefly or give "proof"?

any unsigned integer 'n' have following last k digits: 1 followed (k-1) zeroes: 100...0 note k can 1 in case there no zeroes.

(n - 1) end in format: 0 followed (k-1) 1's: 011...1

n & (n-1) therefore end in 'k' zeroes: 100...0 & 011...1 = 000...0

hence n & (n - 1) eliminate rightmost '1'. each iteration of remove rightmost '1' digit , hence can count number of 1's.

Two