使用struct模塊在python中將2位字符的字符串中的16位ASCII數據解碼為整數時，如何將其解碼？

Question

這是我的代碼。 理想情況下，struct.unpack和encode（'hex'）都將其更改回int應該是正確的嗎？

輸入-

但是，當您給一個.wav文件提供nchannels = 1，samplewidth = 2，framerate = 44100，comptype =“ None”，compname =“ Not Compressed”時，它們是不一樣的

樣品輸出-

-15638 == eac2 == 27330

-15302 == 3ac4 == 15044

-14905 == c7c5 == 18373

-14449 == 8fc7 == 4039

左右手應該相等吧？

import wave
import sys
import struct

audiofile = wave.open(sys.argv[1], 'r')
# reading a file (normal file open)

print audiofile.getparams()
# (nchannels, sampwidth, framerate, nframes, comptype, compname)

for i in range(audiofile.getnframes()):
    frame = audiofile.readframes(1)
    # reading each frame (Here frame is 16 bits [.wav format of each frame])

    print struct.unpack('<h', frame)[0], ' == ',
    # struct.unpack(fmt, string) --- for more info about fmt -> https://docs.python.org/2/library/struct.html
    # If we it is two samples per frame, then we get a tuple with two values -> left and right samples

    value = int(frame.encode('hex'), 16)
    # getting the 16-bit value [each frame is 16 bits]

    if(value > 32767):
        value -= 2**16
    # because wav file format specifies 2's compliment as in even the negative values are there

    print frame.encode('hex') , ' == ', value

audiofile.close()

Answer 1

大端和小端編碼之間的區別。

您的結構是big-endian，而十六進制的轉換是little-endian。