Question

What does h(3231) return for the following function definition?

def h(x):
(m,a) = (1,0)
while m <= x:
(m,a) = (m*2,a+1)
return(a)

Answer 1

Sure. I did mention $x\ne2^n$ for positive innnteger $n$. So, more extended:

$$h(x)=\begin{cases} \lfloor\log_2 x\rfloor &\text{if } x\ne2^n \\ (log_2 x)+1 &\text{if } x=2^n \end{cases}$$

for any arbitrary positive integer $n$.

It can be seen inserting print(m) in the while loop.

So the answer is 12,