我想显示号码
0x40000000
作为
01000000 00000000 00000000 00000000
前导 0 和上面的格式
也许不是最灵活的方式,但是:
>>> num = 0x40000000
>>> bits = bin(num)[2:].zfill(32)
# '01000000000000000000000000000000'
>>> ' '.join(bits[i:i+8] for i in xrange(0, 32, 8))
'01000000 00000000 00000000 00000000'
嗯,因为我的宽带断了,不能早点发帖,但稍微灵活一点的版本......
def fmt_bin(num):
bits = bin(num)[2:]
blocks, rem = divmod(len(bits), 8)
if rem:
blocks +=1
filled = bits.zfill(blocks * 8)
return ' '.join(''.join(el) for el in zip(*[iter(filled)]*8))
这将处理任意大的正数:
def long2str(n):
if n == 0:
return '00000000'
s = []
while n > 0:
s.append('{:08b}'.format(n & 255))
n = n >> 8
return ' '.join(s[::-1])
num = 0x40000000
bignum = 0x4000000040000000
print long2str(num)
print long2str(bignum)
输出:
01000000 00000000 00000000 00000000
01000000 00000000 00000000 00000000 01000000 00000000 00000000 00000000