为什么函数在不指定参数的情况下正常工作？

Question

在下面的代码中，我不明白为什么 download_progress_hook 在从 maybe_download 方法中调用时不传递参数就可以工作。

download_progress_hook 的定义表明必须传递三个参数：count, blockSize, totalSize。 但是，当从maybe_download 调用download_progress_hook 时，没有传递任何参数。为什么不失败？

这里是完整的代码：

url = 'http://commondatastorage.googleapis.com/books1000/'
last_percent_reported = None
data_root = '.' # Change me to store data elsewhere

def download_progress_hook(count, blockSize, totalSize):
  """A hook to report the progress of a download. This is mostly intended for users with
  slow internet connections. Reports every 5% change in download progress.
  """
  global last_percent_reported
  percent = int(count * blockSize * 100 / totalSize)

  if last_percent_reported != percent:
    if percent % 5 == 0:
      sys.stdout.write("%s%%" % percent)
      sys.stdout.flush()
    else:
      sys.stdout.write(".")
      sys.stdout.flush()

    last_percent_reported = percent

def maybe_download(filename, expected_bytes, force=False):
  """Download a file if not present, and make sure it's the right size."""
  dest_filename = os.path.join(data_root, filename)
  if force or not os.path.exists(dest_filename):
    print('Attempting to download:', filename) 
    filename, _ = urlretrieve(url + filename, dest_filename, reporthook=download_progress_hook)
    print('\nDownload Complete!')
  statinfo = os.stat(dest_filename)
  if statinfo.st_size == expected_bytes:
    print('Found and verified', dest_filename)
  else:
    raise Exception(
      'Failed to verify ' + dest_filename + '. Can you get to it with a browser?')
  return dest_filename

train_filename = maybe_download('notMNIST_large.tar.gz', 247336696)
test_filename = maybe_download('notMNIST_small.tar.gz', 8458043)

Answer 1

我得到了一切，但是函数 download_progress_hook 被从函数 maybe_download 中调用的那一点

那是你出错的地方。该函数未被调用。它只是被引用。那里没有(...) 调用表达式。

Python 函数是一流的对象，您可以将它们传递或分配给其他名称：

>>> def foo(bar):
...     return bar + 1
...
>>> foo
<function foo at 0x100e20410>
>>> spam = foo
>>> spam
<function foo at 0x100e20410>
>>> spam(5)
6

这里的spam 是对函数对象foo 的另一个引用。我也可以通过其他名称调用该函数对象。

所以下面的表达式：

urlretrieve(
    url + filename, dest_filename,
    reporthook=download_progress_hook) 

不呼叫 download_progress_hook。它只是将该函数对象提供给urlretrieve() 函数，而该代码将在某处调用download_progress_hook（传入所需的参数）。

来自URLOpener.retrieve documentation（最终处理该钩子）：

如果 reporthook 被给出，它必须是一个接受三个数字参数的函数：一个块号、读入的最大块大小和下载的总大小（如果未知，则为 -1）。它将在开始时和从网络中读取每个数据块之后调用一次。

Answer 2

import urllib.request
import os

class Progress:
def __init__(self):
    self.old_percent = 0

def download_progress_hook(self, count, blockSize, totalSize):
    percent = int(count * blockSize * 100 / totalSize)
    if percent > self.old_percent:
        self.old_percent = percent
        os.system('cls')
        print(percent, '%')
    if percent == 100:
        os.system('cls')
        print('done!')

title = 'title'
url_mp4 = 'https://url'
progress = Progress()
urllib.request.urlretrieve(url_mp4, title + '.mp4', reporthook=progress.download_progress_hook)

Answer 3

我喜欢 ICEBURG 的回答，但 os.system('cls') 不可移植，所以我用 ASCII 制作了一个进度条：

class ProgressBar:
    def __init__(self):
        self.old_percent = 0
        print('_' * 50)

    def download_progress_hook(self, count, blockSize, totalSize):
        percent = int(count * blockSize * 100 / totalSize)
        if percent >= 2 + self.old_percent:
            self.old_percent = percent
            # print(percent, '%')
            print('>', end='')
            sys.stdout.flush()
        if percent == 100:
            print('\ndone!')

它输出：

__________________________________________________
>>>>>>>>>>>>>>>>>>>>>>>>>