CODEWITHSUNDEEP
javabit-shift
Samuele1818
Samuele1818

Reputation: 434

Left circular bit shift

I have to write a function that does a left circular shift of the bits of y positions. For example, if I put: 01011000 and 2 as the y, the function has to return 01100001.

I have tried to use Integer.rotateLeft() but it seems to be useless for this.

Upvotes: 0

Views: 357

Answers (1)

Amir Kirsh
Amir Kirsh

Reputation: 13750

This should work I think:

int rotate_8bits_left(int val, int y) {
    // amend y to the range [0, 7] with this:
    // y = ((y % 8) + 8) % 8;
    // or better, with this:
    y = y & 0x7;
    // do the rotation
    return ((val << y) & 0xFF) | (val >> (8-y));
}

Let's explain the different parts:

// move val left y bits:
(val << y)

Above however keeps the bits that get beyond the 8th bit, so we need to truncate them, so we go with:

// move val left y bits and truncate anything beyond the 8th bit:
(val << y) & 0xFF

Now we need to add the bits that went out, to the beginning. We can calculate the bit that went out to the left by just moving to the right:

// move to the right, 8-y bits
val >> (8-y)

If we now glue together the two parts, we would get the rotation:

int new_val = ((val << y) & 0xFF) | (val >> (8-y));

Now for the first part, we want to handle y that might not be in the range [0, 8]. We can amend y to this range, before using it, with:

y = ((y % 8) + 8) % 8;

The expression above fixes both negative and positive values of y, if y is negative the modulo would return a negative value in the range [-7, -1], then by adding 8 we get back to the positive range. We have to do modulo again, for the case where y was positive and adding 8 to fix the negative case took it back to be above 8. The second modulo fixes this.

But we can achieve the same amendment for y with a more simple approach, by leaving only the 3 first bits that encounter for the range [0, 7], treating 8 as 0, this can be done with the following expression, that works well for both negative and positive values of y:

y = y & 0x7;

Upvotes: 1

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