Puppy Raffle

AI First Flight #1
Beginner FriendlyFoundrySolidityNFT
EXP
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Submission Details
Severity: medium
Valid

M-1 Looping in PuppyRaffle::enterRaffle causing significant increase of gas cost


Description

The PuppyRaffle::enterRaffle function loops through the players array to check for duplicates. However, the longer the PuppyRaffle::players array is, the more checks a new player will have to make. This means the gas costs for players who enter right when the raffle stats will be dramatically lower than those who enter later. Every additional address in the players array, is an additional check the loop will have to make.

Impact

Impact: The gas costs for raffle entrants will greatly increase as more players enter the raffle. Discouraging later users from entering, and causing a rush at the start of a raffle to be one of the first entrants in the queue.

An attacker might make the PuppyRaffle::entrants array so big, that no one else enters, guarenteeing themselves the win.

Proof of Concept

First entered group 100 users -> Gas cost: 23, 990, 822.

Second enter group 100 users -> Gas cost: 88, 937, 355.

Increased on more that 3x cost for second group.

function testDoSEnterRaffle() public {
// Arrange -> Act -> Assert
// First enter group
address[] memory players = new address[](100);
for(uint i = 0; i < 100; i++) {
players[i] = address(uint160(i+1));
}
uint256 gasStartFirstGroup = gasleft();
puppyRaffle.enterRaffle{value: entranceFee * 100}(players);
uint256 gasCostFirstGroup = gasStartFirstGroup - gasleft();
// Second enter group
address[] memory playersSecond = new address[](100);
for(uint i = 0; i < 100; i++) {
playersSecond[i] = address(uint160(101+i));
}
uint256 gasStartSecondGroup = gasleft();
puppyRaffle.enterRaffle{value: entranceFee * 100}(playersSecond);
uint256 gasCostSecondGroup = gasStartSecondGroup - gasleft();
console2.log("Gas cost First: %s", gasCostFirstGroup);
console2.log("Gas cost Second: %s", gasCostSecondGroup);
assert(gasCostSecondGroup > gasCostFirstGroup);
}

Terminal log

Ran 1 test for test/PuppyRaffleTest.t.sol:PuppyRaffleTest[PASS] testDoSEnterRaffle() (gas: 112961697)Logs:Gas cost First: 23990822Gas cost Second: 88937355

Recommended Mitigation

  1. Consider allowing duplicates. Users can make new wallet addresses anyways, so a duplicate check doesn't prevent the same person from entering multiple times, only the same wallet address.

  2. Consider using a mapping to check for duplicates. This would allow constant time lookup of whether a user has already entered.

Updates

Lead Judging Commences

ai-first-flight-judge Lead Judge about 3 hours ago
Submission Judgement Published
Validated
Assigned finding tags:

[M-01] `PuppyRaffle: enterRaffle` Use of gas extensive duplicate check leads to Denial of Service, making subsequent participants to spend much more gas than prev ones to enter

## Description `enterRaffle` function uses gas inefficient duplicate check that causes leads to Denial of Service, making subsequent participants to spend much more gas than previous users to enter. ## Vulnerability Details In the `enterRaffle` function, to check duplicates, it loops through the `players` array. As the `player` array grows, it will make more checks, which leads the later user to pay more gas than the earlier one. More users in the Raffle, more checks a user have to make leads to pay more gas. ## Impact As the arrays grows significantly over time, it will make the function unusable due to block gas limit. This is not a fair approach and lead to bad user experience. ## POC In existing test suit, add this test to see the difference b/w gas for users. once added run `forge test --match-test testEnterRaffleIsGasInefficient -vvvvv` in terminal. you will be able to see logs in terminal. ```solidity function testEnterRaffleIsGasInefficient() public { vm.startPrank(owner); vm.txGasPrice(1); /// First we enter 100 participants uint256 firstBatch = 100; address[] memory firstBatchPlayers = new address[](firstBatch); for(uint256 i = 0; i < firstBatchPlayers; i++) { firstBatch[i] = address(i); } uint256 gasStart = gasleft(); puppyRaffle.enterRaffle{value: entranceFee * firstBatch}(firstBatchPlayers); uint256 gasEnd = gasleft(); uint256 gasUsedForFirstBatch = (gasStart - gasEnd) * txPrice; console.log("Gas cost of the first 100 partipants is:", gasUsedForFirstBatch); /// Now we enter 100 more participants uint256 secondBatch = 200; address[] memory secondBatchPlayers = new address[](secondBatch); for(uint256 i = 100; i < secondBatchPlayers; i++) { secondBatch[i] = address(i); } gasStart = gasleft(); puppyRaffle.enterRaffle{value: entranceFee * secondBatch}(secondBatchPlayers); gasEnd = gasleft(); uint256 gasUsedForSecondBatch = (gasStart - gasEnd) * txPrice; console.log("Gas cost of the next 100 participant is:", gasUsedForSecondBatch); vm.stopPrank(owner); } ``` ## Recommendations Here are some of recommendations, any one of that can be used to mitigate this risk. 1. User a mapping to check duplicates. For this approach you to declare a variable `uint256 raffleID`, that way each raffle will have unique id. Add a mapping from player address to raffle id to keep of users for particular round. ```diff + uint256 public raffleID; + mapping (address => uint256) public usersToRaffleId; . . function enterRaffle(address[] memory newPlayers) public payable { require(msg.value == entranceFee * newPlayers.length, "PuppyRaffle: Must send enough to enter raffle"); for (uint256 i = 0; i < newPlayers.length; i++) { players.push(newPlayers[i]); + usersToRaffleId[newPlayers[i]] = true; } // Check for duplicates + for (uint256 i = 0; i < newPlayers.length; i++){ + require(usersToRaffleId[i] != raffleID, "PuppyRaffle: Already a participant"); - for (uint256 i = 0; i < players.length - 1; i++) { - for (uint256 j = i + 1; j < players.length; j++) { - require(players[i] != players[j], "PuppyRaffle: Duplicate player"); - } } emit RaffleEnter(newPlayers); } . . . function selectWinner() external { //Existing code + raffleID = raffleID + 1; } ``` 2. Allow duplicates participants, As technically you can't stop people participants more than once. As players can use new address to enter. ```solidity function enterRaffle(address[] memory newPlayers) public payable { require(msg.value == entranceFee * newPlayers.length, "PuppyRaffle: Must send enough to enter raffle"); for (uint256 i = 0; i < newPlayers.length; i++) { players.push(newPlayers[i]); } emit RaffleEnter(newPlayers); } ```

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