# Double-nested loop with bitwise operation

I have this little exercise:

for ( i = 0; i < 2 * n; i += 2 )
for ( j = 1; j <= n; j <<= 1 )
if ( j & i )
foo ();


(j <<= 1 means j = (j << 1), where << is the bitwise left shift operator. & is the bitwise and operator.)

I am supposed to determine how many times will the foo function be called for some n. The result should be both an exact number (or the most accurate approximation possible) and asymptotic (like O(n)).

## migrated from stackoverflow.comSep 22 '12 at 22:56

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• Suggestion: you could try compiling and running the code with a few different values of n, just to get a feel for how the answers should look. – Paul R Jan 12 '12 at 17:29
• It's analysis so I should use just pencil and paper. – Machta Jan 12 '12 at 17:35
• Running the code is (a) instructive and (b) a useful reality-check against any theoretical answers that you come up with – Paul R Jan 12 '12 at 22:16

The number of foo() calls is the number of 1 bits in all numbers from 0 to $n$ (or all even numbers between $0$ and $2n$ - same thing). Asymptotically, it's $O(n \log n)$, because the number of 1 bits is around $\log_2(n)/2$.
I don't have an exact function. It won't be a smooth function - for example, $f(1023)-f(1022)=10$, but $f(1024)-f(1023)=1.$