tornadocash/fixed-merkle-tree
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Fix lint erros

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This commit is contained in:
Rodrigo Q. Saramago 2022-07-15 10:33:37 +02:00
parent 8ce9847652
commit f9d7765305
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GPG Key ID: 9B36B2525704A359
2 changed files with 8 additions and 8 deletions
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14
src/BaseTree.ts
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@ -141,7 +141,7 @@ export class BaseTree {
* @returns {number[]} the new list of indices
*/
static nextLayerMultiPathIndices(indices: number[]): number[] {
let nextIndices: Set<number> = new Set()
const nextIndices: Set<number> = new Set()
for (let i = 0; i < indices.length; i++) {
nextIndices.add(indices[i] >> 1)
}
@ -158,10 +158,10 @@ export class BaseTree {
let layerIndices = indices
for (let level = 0; level < this.levels; level++) {
// find whether there is a neighbor idx that is not in layerIndices
let proofElements = layerIndices.reduce((elements, idx) => {
const proofElements = layerIndices.reduce((elements, idx) => {
const leafIndex = idx ^ 1
if (!layerIndices.includes(leafIndex)) {
if (leafIndex < this._layers[level].length) {
if (leafIndex < this._layers[level].length) {
elements.push(this._layers[level][leafIndex])
} else {
elements.push(this._zeros[level])
@ -199,14 +199,14 @@ export class BaseTree {
): boolean {
let layerElements: Element[] = leaves
let layerIndices: number[] = leafIndices
let proofElements: Element[] = pathElements
let layerProofs: Element[] = []
const proofElements: Element[] = pathElements
const layerProofs: Element[] = []
for (let level = 0; level < levels; level++) {
for (let i = 0; i < layerIndices.length; i++) {
let layerHash: string
let elIndex = layerIndices[i]
let leafIndex = elIndex ^ 1
const elIndex = layerIndices[i]
const leafIndex = elIndex ^ 1
if (layerIndices.includes(leafIndex)) {
if (elIndex % 2 === 0) {
layerHash = hashFn(layerElements[0], layerElements[1])

2
src/FixedMerkleTree.ts
View File

@ -70,7 +70,7 @@ export default class MerkleTree extends BaseTree {
}
multiProof(elements: Element[]): MultiProofPath {
let indexes = []
const indexes = []
for (let i = 0; i < elements.length; i++) {
indexes.push(this.indexOf(elements[i]))
}