如何转换从文本文件中读取的整数并存储为具有16位整数的二进制文件？

Question

尝试从文本文件中读取十进制值转换为16位二进制文​​件并转换为二进制文件。

示例输入文件

120
300
-250
13
-120

码：

def decimaltoBinary(filename,writefile):
    file = filename
    print(file)
    file_write = open(writefile,'wb')
    file_read = open(file, 'rb')
    for line in file_read:
        value = int(line)
        if value < 0:
            binary_value = bin((2**16) - abs(value))[2:].zfill(16)
            file_write.write(binary_value + "\n")
        else:
            binary_value = bin(int(value))[2:].zfill(16)
            file_write.write(binary_value + "\n")
    file_write.close()

decimaltoBinary(input_file.text,output_file.bin)

希望将转换后的十进制值写入二进制文件..非常感谢任何帮助

Answer 1

您可以使用the struct module：

data = [120,300,-250,13,-120] # you seem to have the reading part covered already
                              # using a list as data input for demo purposes
import struct

with open("f.bin","wb") as f: 
    for d in data:
        f.write(struct.pack('h', d)) # 2 byte integer aka short

with open("f.bin","rb") as f:
    print(f.read())  # b'x\x00,\x01\x06\xff\r\x00\x88\xff'

您只需指定'h'即可获得短（2字节整数）打包。

对于奇怪的打印输出blame python - 它用较短的正常字符替换“已知”\xXX代码 - f.e. ',' => 0x2c或\r => \x0d等。