If you have a byte array of length l, how can you pad it by any number say x?

Normal obvious way is below.

```
if (l%x == 0) return 0
else return (x - (l%x))
```

However,

**if x is a power of 2**, a faster approach is to get the two's complement (2C) of length l and capture the least significant bits.

i.e.

```
(l*-1)&(x-1)
```

```
```

l=5, x=4

00000101 //5

11111011 //-5

00000011 //4-1=3

00000011 //-5&3 = 3

So, for l=5 you need 3 bytes to pad.

l=7, x=4

00000111 //7

11111001 //-7

00000011 //4-1=3

00000001 //-7&3 = 1

So, for l=7 you need 1 bytes to pad.

It only works for x which is a power of 2. For other x, you have to go by the normal obvious method.

Check if x is a power of 2 this way:

```
bool isPowerOf2(den):
return (den & (den-1)) == 0
```

[Hat tip to JvE]

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