First Flight #18: T-Swap

First Flight #18
Beginner FriendlyDeFiFoundry
100 EXP
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Submission Details
Severity: high
Valid

In `TSwapPool::_swap` the extra tokens given to users after every `swapCount` breaks the protocol invariant of `x * y = k`

In TSwapPool::_swap the extra tokens given to users after every swapCount breaks the protocol invariant of x * y = k

Description: The protocol follows a strict invariant of x * y = k. Where:

  • x: The balance of the pool token

  • y: The balance of WETH

  • k: The constant product of the two balances

This means, that whenever the balances change in the protocol, the ratio between the two amounts should remain constant, hence the k. However, this is broken due to the extra incentive in the _swap function. Meaning that over time the protocol funds will be drained.

The issue is found in the next code.

swap_count++;
if (swap_count >= SWAP_COUNT_MAX) {
swap_count = 0;
outputToken.safeTransfer(msg.sender, 1_000_000_000_000_000_000);
}

Impact: A user could drain the protocol by doing a lot of swaps and collecting the extra incentive given out by the protocol.

Proof of Concept:

  1. A user swaps 10 times, and collects the extra incentive of 1_000_000_000_000_000_000 tokens

  2. That user continues to swap untill all the protocol funds are drained

Proof Of Code

Place the following into TSwapPool.t.sol.

function testInvariantBroken() public {
vm.startPrank(liquidityProvider);
weth.approve(address(pool), 100e18);
poolToken.approve(address(pool), 100e18);
pool.deposit(100e18, 100e18, 100e18, uint64(block.timestamp));
vm.stopPrank();
uint256 outputWeth = 1e17;
vm.startPrank(user);
poolToken.approve(address(pool), type(uint256).max);
poolToken.mint(user, 100e18);
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
int256 startingY = int256(weth.balanceOf(address(pool)));
int256 expectedDeltaY = int256(-1) * int256(outputWeth);
pool.swapExactOutput(poolToken, weth, outputWeth, uint64(block.timestamp));
vm.stopPrank();
uint256 endingY = weth.balanceOf(address(pool));
int256 actualDeltaY = int256(endingY) - int256(startingY);
assertEq(actualDeltaY, expectedDeltaY);
}

Recommended Mitigation: Remove the extra incentive mechanism. If you want to keep this in, we should account for the change in the x * y = k protocol invariant. Or, we should set aside tokens in the same way we do with fees.

- swap_count++;
- // Fee-on-transfer
- if (swap_count >= SWAP_COUNT_MAX) {
- swap_count = 0;
- outputToken.safeTransfer(msg.sender, 1_000_000_000_000_000_000);
- }
Updates

Appeal created

inallhonesty Lead Judge 12 months ago
Submission Judgement Published
Validated
Assigned finding tags:

In `TSwapPool::_swap` the extra tokens given to users after every swapCount breaks the protocol invariant of x * y = k

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