我正在读取python中的二进制文件,并且文件格式的文档说:
标记(二进制)表示
1 nnn nnnn表示要跟随一个数据字节可以重复的nnn nnnn(最多127个)次。
0 nnn nnnn表示图像有nnn nnnn个字节后续数据(最大127个字节),并且没有重复。
n 000 0000行尾字段。指示行尾记录。 n的值可以为零或一。请注意,行尾字段是必填字段,它反映在行记录的长度中上面提到的字段。
[读取文件时,我希望我所在的字节返回1 nnn nnnn
,其中nnn nnnn
部分应为50。
我已经可以使用以下方法做到这一点:
flag = byte >> 7
numbytes = int(bin(byte)[3:], 2)
但是计算numbytes感觉很便宜。
我可以做更多的数学运算来完成对numbytes的计算吗?
您将如何处理?
mask = 0b01111111
byte_from_file = 0b10101010
value = mask & byte_from_file
print bin(value)
>> 0b101010
print value
>> 42
在进行位掩码时,我发现二进制数字比十六进制更容易理解。编辑:您的用例的稍微完整的示例:
LEADING_BIT_MASK = 0b10000000 VALUE_MASK = 0b01111111 values = [0b10101010, 0b01010101, 0b0000000, 0b10000000] for v in values: value = v & VALUE_MASK has_leading_bit = v & LEADING_BIT_MASK if value == 0: print "EOL" elif has_leading_bit: print "leading one", value elif not has_leading_bit: print "leading zero", value
x = 10 # 1010 in binary
if x & 0b10: # explicitly: x & 0b0010 != 0
print('First bit is set')
要检查是否设置了第n ^位,使用2的幂或更好的移位
def is_set(x, n): return x & 2 ** n != 0 # a more bitwise- and performance-friendly version: return x & 1 << n != 0 is_set(10, 1) # 1 i.e. first bit - as the count starts at 0-th bit >>> True
if (byte & 0x80) != 0:
num_bytes = byte & 0x7F
>>> x = 154 #just an example
>>> flag = x >> 1
>>> flag
1
>>> nb = x & 127
>>> nb
26
def GetVal(b):
# mask off the most significant bit, see if it's set
flag = b & 0x80 == 0x80
# then look at the lower 7 bits in the byte.
count = b & 0x7f
# return a tuple indicating the state of the high bit, and the
# remaining integer value without the high bit.
return (flag, count)
>>> testVal = 50 + 0x80
>>> GetVal(testVal)
(True, 50)
class ControlWord(object):
"""Helper class to deal with control words.
Bit setting and checking methods are implemented.
"""
def __init__(self, value = 0):
self.value = int(value)
def set_bit(self, bit):
self.value |= bit
def check_bit(self, bit):
return self.value & bit != 0
def clear_bit(self, bit):
self.value &= ~bit