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                ThinkChat2.0新版上線,更智能更精彩,支持會話、畫圖、視頻、閱讀、搜索等,送10W Token,即刻開啟你的AI之旅 廣告
                struct?模塊為使用變長的二進制記錄格式提供了?pack()?和?unpack()?函數。下面的示例演示了在不使用?zipfile?模塊的情況下如何迭代一個 ZIP 文件的頭信息。壓縮碼?"H"?和?"I"?分別表示2和4字節無符號數字,?"?表明它們都是標準大小并且按照 little-endian 字節排序。 ~~~ import struct with open('myfile.zip', 'rb') as f: data = f.read() start = 0 for i in range(3): # show the first 3 file headers start += 14 fields = struct.unpack(', data[start:start+16]) crc32, comp_size, uncomp_size, filenamesize, extra_size = fields start += 16 filename = data[start:start+filenamesize] start += filenamesize extra = data[start:start+extra_size] print(filename, hex(crc32), comp_size, uncomp_size) start += extra_size + comp_size # skip to the next header ~~~
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