[FIXED] How to remove the nth hexadecimal digit of a integer number without using Strings?

Issue

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?

Solution

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));
}

Answered By - Falk Hüffner

Answer Checked By - Senaida (PHPFixing Volunteer)