Consider a hexadecimal integer value such as `n = 0x12345`

, how to get `0x1235`

as result by doing `remove(n, 3)`

(big endian)?

For the inputs above I think this can be achieved by performing some bitwising steps:

`partA`

= extract the part from index`0`

to`targetIndex - 1`

(should return`0x123`

);`partB`

= extract the part from`targetIndex + 1`

to`length(value) - 1`

(`0x5`

);- result, then, can be expressed by
`((partA << length(partB) | partB)`

, giving the`0x1235`

result.

However I’m still confused in how to implement it, once each hex digit occupies 4 spaces. Also, I don’t know a good way to retrieve the length of the numbers.

This can be easily done with strings however I need to use this in a context of thousands of iterations and don’t think Strings is a good idea to choose.

So, what is a good way to this removing without Strings?

## Answer

Similar to the idea you describe, this can be done by creating a mask for both the upper and the lower part, shifting the upper part, and then reassembling.

int remove(int x, int i) { // create a mask covering the highest 1-bit and all lower bits int m = x; m |= (m >>> 1); m |= (m >>> 2); m |= (m >>> 4); m |= (m >>> 8); m |= (m >>> 16); // clamp to 4-bit boundary int l = m & 0x11111110; m = l - (l >>> 4); // shift to select relevant position m >>>= 4 * i; // assemble result return ((x & ~(m << 4)) >>> 4) | (x & m); }

where “>>>” is an unsigned shift.

As a note, if 0 indicates the highest hex digit in a 32-bit word independent of the input, this is much simpler:

int remove(int x, int i) { int m = 0xffffffff >>> (4*i); return ((x & ~m) >>> 4) | (x & (m >>> 4)); }