javascript - 如何修改以下回调函数以使其返回 promise ？

Question

此函数尝试加载图像，如果图像未加载，则再次调用该函数:

function loadImage (url = '', callback = () => {}) {
  utils.loadImage(url, () => {
    callback()
  }, () => {
    loadImage(url, callback)
  })
}

我保证它会回来，所以我写道:

function loadImage (url = '', callback = () => {}) {
  return new Promise((resolve, reject) => {
    utils.loadImage(url, () => {
      // what to do with resolve and reject
      callback()
    }, () => {
      loadImage(url, callback)
    })
  })
}

但我不确定将 resolve 和 reject 放在哪里......

Answer 1

const loadImage = url => new Promise(resolve => {
  utils.loadImage(url, resolve, () => {
    loadImage(url).then(resolve)
  })
})

编辑:由于 Bergi 的评论，版本更好:

const loadImage = url => new Promise(resolve => {
  utils.loadImage(url, resolve, () => resolve(loadImage(url)))
})