totalFees Overflow Causes Silent Loss of All Fee Revenue
Severity: Medium
Description
The protocol accumulates a 20% fee from each raffle round into totalFees, which the owner withdraws via withdrawFees().
totalFees is declared as uint64 with a maximum value of approximately 18.44 ETH. When the fee from a raffle round exceeds this, the unsafe cast uint64(fee)
silently wraps to near-zero. In Solidity 0.7.6 there is no automatic overflow protection, so no revert occurs and no event is emitted.
// @> totalFees declared as uint64 — max ~18.44 ETH
uint64 public totalFees = 0;
function selectWinner() external {
uint256 fee = (totalAmountCollected * 20) / 100;
// @> uint256 fee silently truncated to uint64
// @> 93 players @ 1 ETH = 18.6 ETH fee -> overflows, stored as 0.15 ETH
totalFees = totalFees + uint64(fee);
}
Risk
Likelihood:
The overflow triggers with as few as 93 players at 1 ETH entrance fee — a realistic participation level for any active raffle round.
The overflow is completely silent — no revert, no event, no indication that fee accounting is broken.
Impact:
The protocol owner permanently loses the majority of fee revenue with no recovery mechanism.
withdrawFees() succeeds but sends only the small wrapped value instead of actual fees earned.
Proof of Concept
With 100 players at 1 ETH each, the 20 ETH fee exceeds uint64 max and wraps to approximately 1.55 ETH. The owner calls withdrawFees() successfully but receives
only a fraction of the actual fees:
function test_TotalFees_Overflow() public {
uint256 numPlayers = 100;
address[] memory players = new address;
for (uint256 i = 0; i < numPlayers; i++)
players[i] = address(uint160(100 + i));
vm.deal(address(this), entranceFee * numPlayers);
raffle.enterRaffle{value: entranceFee * numPlayers}(players);
vm.warp(block.timestamp + duration + 1);
raffle.selectWinner();
uint64 stored = raffle.totalFees();
uint256 expected = (100 * entranceFee * 20) / 100;
assertLt(uint256(stored), expected);
}
Recommended Mitigation
Change totalFees to uint256 to match the type of the fee calculation and eliminate the unsafe cast entirely:
uint64 public totalFees = 0;
uint256 public totalFees = 0;
totalFees = totalFees + uint64(fee);
totalFees = totalFees + fee;
## Description ## Vulnerability Details The type conversion from uint256 to uint64 in the expression 'totalFees = totalFees + uint64(fee)' may potentially cause overflow problems if the 'fee' exceeds the maximum value that a uint64 can accommodate (2^64 - 1). ```javascript totalFees = totalFees + uint64(fee); ``` ## POC <details> <summary>Code</summary> ```javascript function testOverflow() public { uint256 initialBalance = address(puppyRaffle).balance; // This value is greater than the maximum value a uint64 can hold uint256 fee = 2**64; // Send ether to the contract (bool success, ) = address(puppyRaffle).call{value: fee}(""); assertTrue(success); uint256 finalBalance = address(puppyRaffle).balance; // Check if the contract's balance increased by the expected amount assertEq(finalBalance, initialBalance + fee); } ``` </details> In this test, assertTrue(success) checks if the ether was successfully sent to the contract, and assertEq(finalBalance, initialBalance + fee) checks if the contract's balance increased by the expected amount. If the balance didn't increase as expected, it could indicate an overflow. ## Impact This could consequently lead to inaccuracies in the computation of 'totalFees'. ## Recommendations To resolve this issue, you should change the data type of `totalFees` from `uint64` to `uint256`. This will prevent any potential overflow issues, as `uint256` can accommodate much larger numbers than `uint64`. Here's how you can do it: Change the declaration of `totalFees` from: ```javascript uint64 public totalFees = 0; ``` to: ```jasvascript uint256 public totalFees = 0; ``` And update the line where `totalFees` is updated from: ```diff - totalFees = totalFees + uint64(fee); + totalFees = totalFees + fee; ``` This way, you ensure that the data types are consistent and can handle the range of values that your contract may encounter.
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