这是我的密码。理想情况下两者都有解包结构和encode('hex')并将其改回int应该是相同的对吗?
输入-
But, they are not the same in this case when you give a .wav file with nchannels = 1, samplewidth = 2, framerate = 44100, comptype = "None", compname = "Not compressed"
样本输出-
-15638 == eac2 == 27330
-15302 == 3ac4 == 15044
-14905 == c7c5 == 18373
-14449 == 8fc7 == 4039
The left and right hand-side should be equal right?
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()
区别在于大端和小端编码。在
你的结构是big-endian,而hex的转换是little-endian。在
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