Python，阅读Hex Little Endian用它进行数学数，并将结果覆盖为旧位置

Question

我是新手编程，我需要帮助我有一个类似的十六进制文件：

43 52 53 00 00 00 00 00 00 00 01 01 01 30 00 00 00 00

10 87 01 00 13 00 00 00 10 00 00 00 00 00 00 00 00 00

40 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

我需要python代码，让我阅读小endian" 10 87 01"做数学

喜欢10 87 01 40 01 = 50 88 01

43 52 53 00 00 00 00 00 00 00 01 01 01 30 00 00 00 00

50 88 01 00 13 00 00 00 10 00 00 00 00 00 00 00 00 00

40 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

希望它的清晰

Answer 1

您可以使用struct库来处理小恩迪安P>

对于您的特定任务，我不知道我是否正确理解，因为您没有指定二进制文件中有哪种数据...让我们假设我们已经签署了32位整数，我的代码将成为这样的事情：

import struct
# we are assuming you have 32 bit integers on your file
block_size = 4
filename = "prova.bin"

# function to do "some math... :)"
def do_some_math(my_hex_value):
    return my_hex_value + 1

# open and read the whole file
with open(filename, "r+b") as f:
    my_byte = f.read(block_size)
    while(len(my_byte) == block_size):
        # unpack the 4 bytes value read from file
        # "<" stands for "little endian"
        # "i" stands for "integer"
        # more info on the struct library in the official doc
        my_hex_value = struct.unpack_from("<i", my_byte)[0]
        print("Before math = " + str(my_hex_value))
        # let's do some math
        my_hex_value = do_some_math(my_hex_value)
        print("After math = " + str(my_hex_value))
        # let's repack the hex back
        my_byte = struct.pack("<i", my_hex_value)
        # let's reposition the file pointer so as to overwrite
        # the bytes we have previously read 
        f.seek(f.tell() - block_size)
        # let's override the old bytes
        f.write(my_byte)
        # let's read another chunk to repeat till the eof
        my_byte = f.read(block_size)

希望这会有所帮助
最好的
dave

Answer 2

假设您的数学功能可以计算新模式，则可以使用这样的函数：

def replace_pattern(old_pattern, new_pattern, occurrence):
with open('input_file.txt','r') as myfile:
    myline=""
    for line in myfile:
        myline += line
if occurrence == 0: # assume 0 is used to indicate all occurrences must be replaced
    if myline.find(old_pattern) == -1:
        print('pattern not found, exit')
        return
    else:
        newline = myline.replace(old_pattern, new_pattern)
else: #a particular occurrence has to be updated
    idx = 0
    offset=0
    nbmatch = 0
    while idx != -1:
        idx = myline.find(old_pattern, offset)
        if idx != -1:
            offset = idx+1
            nbmatch += 1
        if nbmatch == occurrence:
            # the index of the target occurrence has been reached
            break
    if nbmatch == 0:
        print('problem, at least one occurrence expected')
        return
    elif nbmatch == 1:
        print('problem, more than one occurrence expected, replace anyway')
        newline = myline.replace(old_pattern, new_pattern)
    else:
        # further processing on a part of the line
        sameline = myline[:idx]
        diffline = myline[idx:]
        # work on diffline substring
        diffline = diffline.replace(old_pattern,new_pattern,1)
        # rebuild line
        newline = sameline+diffline
with open('input_file.txt','w') as myfile:
    myfile.write(newline)

我可能没有被优化，但应该按预期工作