How to efficiently replace a contiguous set of bits

Let's say I have a binary number 1001100, and I want to substitute it with 1011 at the section [2, 6).

It would look something like this:

Binary: 1001100
Sub:    --1011-
Result: 1010110

I know it's possible to repeatedly set a single bit multiple times using this code:
number |= 1UL << n;

But I was wondering if there was a way to generally achieve this, in a more efficient way.

Solution

Lets consider a more general problem: you have a binary number a, and you want to replace its bits by bits of another number b, only when the corresponding bit of third number m is one.

For your specific problem:

a = 1001100
b = xx1011x
m = 0011110

but there is no need to have m correspond to a single range.

A pseudo code to compute the result r would be

for all i in 0 to 6
  if m[i]=1
     r[i]=a[i]
  else
    r[i]=b[i]

or equivalently

for all i in 0 to 6
  r[i] = (a[i] and (m[i]=1)) or (b[i] and (m[i]=0))

We can easily deduce from this that the loop is useless and that the expression

r = (a & m) | (b & ~m)

gives the correct result.

Note that if you want to have a replacement on a specific range k to l (1 to 5 for instance in your case), the mask will be (2^l - 1) - (2^k - 1) = 2^l - 2^k

So here is corresponding C code

unsigned int replace_bit(unsigned int a, unsigned int b, int k, int l){
  unsigned int mask=(1<<l) - (1<<k)
  return  (a & mask) | (b & ~mask) ;
}