输出一定范围unicode对应符号

#本程序没有考虑对0x的处理，请勿输入，直接输入16进制位即可

begin = input("起始点：")

end = input("结束点：")

b_int0 = 0

e_int0 = 0

b_len = len(begin)

e_len = len(end)

dit0 = ['a','b','c','d','e','f']

dit1 = ['A','B','C','D','E','F']

dit = dit0[:] + dit1[:]

if begin>end or begin == end:                           #筛选条件，待补充，有漏洞，如‘FF’与‘1FF’

    print("error!!")

else:

     for i in range(b_len):                                 #按照0x56 = 5*16+6的思路进行逐位运算

         if begin[i] in dit:

             for j in range(0,6,1):

                 if begin[i] == dit0[j]:                    #此处分两词典只是单纯适应原程序

                     b = 10 + j

                 if begin[i] == dit1[j]:                   #16进制有A~F，应分开筛选

                     b = 10 + j

         elif int (begin[i]) <= 9:

                 b = int (begin[i])

         b_len = b_len-1                                    #此处是为了降低运算位，从效果上看并不影响前面的range(b_len)

         b = b*16**(b_len)

         b_int0 += b

     for i in range(e_len):

         if end[i] in dit:

             for j in range(0,6,1):

                 if end[i] == dit0[j]:

                     e = 10 + j

                 if end[i] == dit1[j]:

                     e = 10 + j

         elif int (end[i]) <= 9:

                 e = int (end[i])

         e_len = e_len-1

         e = e*16**(e_len)

         e_int0 += e

for i in range(b_int0,e_int0+1,1):

    print(chr(i))       

附：下列转载自：https://blog.csdn.net/gywtzh0889/article/details/71083459