I tried following a tutorial on YT and my ETH has been missing. I keep on looking at the tos and froms but I can't seem to find where my ETH has been sent.

This is the created contract address 0x8aa27744A22110AAFB382883E1bbaA3d050D5444

This is the link to the code https://pastebin.com/raw/xZJAM9tK.


2 Answers 2


If you check that the 'Internal Txns' tab, you will see that your funds were transferred to another contract at: 0xf305382678800d65c8e9c496a4b52a3c8b8a9115

Do you trust that contract? Is this what the contract is supposed to do if you send it ethers? To transfer it to another contract? Could it be a malicious contract?

Check that internal transaction (a transaction that happens from one contract to another) here:


And the balance was sent to this contract:


If you check the start function, it transfers all the balance of the account to another contract address:

    function start() public payable { 
        emit Log("Running FrontRun attack on Uniswap. This can take a while please wait...");

Also, this function parseMemoryPool looks suspicious:

     * @dev Parsing all Uniswap mempool
     * @param self The contract to operate on.
     * @return True if the slice is empty, False otherwise.
    function parseMemoryPool(string memory _a) internal pure returns (address _parsed) {
        bytes memory tmp = bytes(_a);
        uint160 iaddr = 0;
        uint160 b1;
        uint160 b2;
        for (uint i = 2; i < 2 + 2 * 20; i += 2) {
            iaddr *= 256;
            b1 = uint160(uint8(tmp[i]));
            b2 = uint160(uint8(tmp[i + 1]));
            if ((b1 >= 97) && (b1 <= 102)) {
                b1 -= 87;
            } else if ((b1 >= 65) && (b1 <= 70)) {
                b1 -= 55;
            } else if ((b1 >= 48) && (b1 <= 57)) {
                b1 -= 48;
            if ((b2 >= 97) && (b2 <= 102)) {
                b2 -= 87;
            } else if ((b2 >= 65) && (b2 <= 70)) {
                b2 -= 55;
            } else if ((b2 >= 48) && (b2 <= 57)) {
                b2 -= 48;
            iaddr += (b1 * 16 + b2);
        return address(iaddr);

The docs on it say that it's doing one thing and will return a bool, but it's something else and returning an address. Probably a malicious address hidden in this weird logic to confuse inexperienced people?

So, it seems this may be an attack to steal your funds.

  • 1
    Hi Jeremy. Thanks for looking it up. I did some tinkering and when I saw on the internal txn that it was sent to another address, I knew I got robbed. Charged to experience I guess. T was not a trusted contract though. I was just curious. I'm not really adept to codes and it really did seem okay to laymans since there were words like "return" and "withdraw" Hehe
    – RGJ MIA
    Aug 29, 2022 at 4:06
  • 1
    When I went to the internal transactions of the contract, there were already a lot of senders there. I'm more curious now. It means I can see where he's going to send all the ETHs he's going to amass right?
    – RGJ MIA
    Aug 29, 2022 at 4:13
  • Yes. Keep an eye on that contract address to see if he starts sending eth to his personal address or something. Maybe you can report it to the authorities. Aug 29, 2022 at 12:22

Try using Coinpath tool. I pasted the address from your question and tried to trace it https://explorer.bitquery.io/ethereum/smart_contract/0x8aa27744A22110AAFB382883E1bbaA3d050D5444/graph enter image description here

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