#C 35 bytes

N;f(n){for(N=n;n&n-1;)n&=n-1;N-=n;}

Try it online!

n&=n-1

Does the opposite of the problem statement, it clears the least significant bit.
I have a vague memory of there being a similar trick for most significant bit from seeing it on coding game but may remember wrong.

Recursive function, (削除) 45 (削除ここまで) 43 bytes

N;g(n){n&=(N=n&~-n)?g(N):n;}f(n){N=n-g(n);}

Try it online!

cleblancs code shortened to 34 bytes

i=1;f(n){for(;n/i/2;i*=2);n-=n^i;}

Try it online!

28 bytes, possibly cheating , using pointer instead of return

f(*n){*n^=1<<(int)log2(*n);}

Try it on ideone! , doesn't work on tio.run

C 35 bytes

N;f(n){for(N=n;n&n-1;)n&=n-1;N-=n;}

Try it online!

n&=n-1

Does the opposite of the problem statement, it clears the least significant bit.
I have a vague memory of there being a similar trick for most significant bit from seeing it on coding game but may remember wrong.

Recursive function, (削除) 45 (削除ここまで) 43 bytes

N;g(n){n&=(N=n&~-n)?g(N):n;}f(n){N=n-g(n);}

Try it online!

cleblancs code shortened to 34 bytes

i=1;f(n){for(;n/i/2;i*=2);n-=n^i;}

Try it online!

28 bytes, possibly cheating , using pointer instead of return

f(*n){*n^=1<<(int)log2(*n);}

Try it on ideone! , doesn't work on tio.run

#C 35 bytes

N;f(n){for(N=n;n&n-1;)n&=n-1;N-=n;}

Try it online!

n&=n-1

Does the opposite of the problem statement, it clears the least significant bit.
I have a vague memory of there being a similar trick for most significant bit from seeing it on coding game but may remember wrong.

Recursive function, (削除) 45 (削除ここまで) 43 bytes

N;g(n){n&=(N=n&~-n)?g(N):n;}f(n){N=n-g(n);}

Try it online!

cleblancs code shortened to 34 bytes

i=1;f(n){for(;n/i/2;i*=2);n-=n^i;}

Try it online!

#C 35 bytes

N;f(n){for(N=n;n&n-1;)n&=n-1;N-=n;}

Try it online!

n&=n-1

Does the opposite of the problem statement, it clears the least significant bit.
I have a vague memory of there being a similar trick for most significant bit from seeing it on coding game but may remember wrong.

Recursive function, (削除) 45 (削除ここまで) 43 bytes

N;g(n){n&=(N=n&~-n)?g(N):n;}f(n){N=n-g(n);}

Try it online!

cleblancs code shortened to 34 bytes

i=1;f(n){for(;n/i/2;i*=2);n-=n^i;}

Try it online!

28 bytes, possibly cheating , using pointer instead of return

f(*n){*n^=1<<(int)log2(*n);}

Try it on ideone! , doesn't work on tio.run

#C 35 bytes

N;f(n){for(N=n;n&n-1;)n&=n-1;N-=n;}

Try it online!

n&=n-1

Does the opposite of the problem statement, it clears the least significant bit.
I have a vague memory of there being a similar trick for most significant bit from seeing it on coding game but may remember wrong.

Recursive function, 45 bytes

N;g(n){n=n&(N=n&~-n)?g(N):n;}f(n){n-=n^g(n);}

Try it online!

cleblancs code shortened to 34 bytes

i=1;f(n){for(;n/i/2;i*=2);n-=n^i;}

Try it online!

#C 35 bytes

N;f(n){for(N=n;n&n-1;)n&=n-1;N-=n;}

Try it online!

n&=n-1

Does the opposite of the problem statement, it clears the least significant bit.
I have a vague memory of there being a similar trick for most significant bit from seeing it on coding game but may remember wrong.

Recursive function, (削除) 45 (削除ここまで) 43 bytes

N;g(n){n&=(N=n&~-n)?g(N):n;}f(n){N=n-g(n);}

Try it online!

cleblancs code shortened to 34 bytes

i=1;f(n){for(;n/i/2;i*=2);n-=n^i;}

Try it online!