Byte32 to uint and then back to byte32

Question

I have to bitpack two uint128 values into a uint256, then get back the two original uint128s.

To encode the values I'm using:

function encode256(uint128 a, uint128 b) pure public returns (uint256) {
    return uint256(bytes32(abi.encodePacked(a, b)));
}

For: a=1, b=16 we get a=0x00000000000000000000000000000001 b=0x00000000000000000000000000000010 which leads to 0x0000000000000000000000000000000100000000000000000000000000000010 which cast to uint256 is 340282366920938463463374607431768211472.

The issue is decoding it back to the originals. To unravel the encoded uint256 back to the two uint128 I wrote the following code:

let hex = numberToHex("340282366920938463463374607431768211472") //1 /16
hex = padRight(hex, 64)
hex = hex.replace('0x','') 
const half = Math.ceil(hex.length / 2);    
const firstHalf = hex.slice(0, half)
const secondHalf = hex.slice(half)
console.log(firstHalf)
console.log(secondHalf)

But the two numbers coming out are a=10000000000000000000000000000001, b=00000000000000000000000000000000.

I'm not sure how the encoding is happening and wondering where I am going wrong in the encoding or decoding process?

Answer 1

You are padding right. You need to actually pad left:

hex = padLeft(hex, 64)

If you convert 340282366920938463463374607431768211472 to hex, it would be: 100000000000000000000000000000010. If we left pad it, it will be: 0000000000000000000000000000000100000000000000000000000000000010.

When slicing the first part, it will be: 00000000000000000000000000000001. When slicing the second part, it will be: 00000000000000000000000000000010.

Which is what you want.