如何拆分一个ASCII字符的十六进制字节 [英] How to split hex byte of an ASCII character
问题描述
基本上什么我想要做的就是
有关如:'A'六角equivalant是 0x61
,我可以拆分 61
到 6
和 1
,并将其保存为'6'
和 1
?
一个缓冲区这样接收数据:
rx_dataframe.data [0]为H'00,'。'// H'是十六进制equivalant和''是ASCII值
rx_dataframe.data [0]是H'31,'1'
rx_dataframe.data [0]是H'32,'2'
rx_dataframe.data [0]是H'33,'3'
我需要转换进制数值0x00,0x31,0x32,0x33
中char值 0,0,3 ','1','3','2';'3','3'
并将它们存储在 tx_buff_data []的位置;
我希望我的 tx_buff_data code>这个样子
tx_buff_data [0]有H'30,0
tx_buff_data [1]具有H'30,'0'
tx_buff_data [2]具有H'33,'3'
tx_buff_data [3]具有H'31,'1'
tx_buff_data [4]有H'33,'3'
tx_buff_data [5]具有H'32,'2'
tx_buff_data [6]具有H'33,'3'
tx_buff_data [7]具有H'33,'3'
您可以将每个字节分为使用逐两个半字节(4位数量)和+变化:
unsigned char型LO =字节和放大器;为0x0F;
unsigned char型HI =(字节>> 4)及为0x0F;
然后,您可以各占一半转换成由阵列查找一个十六进制字符(因为字符串本身就是一些字符数组):
字符loChar =0123456789ABCDEF[LO]
焦炭hiChar =0123456789ABCDEF[喜]
What basically i want to do is
For eg: 'a' hex equivalant is 0x61
, can i split61
in to 6
and 1
and store them as '6'
and '1'
?
A buffer is receiving data like this:
rx_dataframe.data[0] is H'00,'.'// H' is Hex equivalant and '' is ASCII value
rx_dataframe.data[0] is H'31,'1'
rx_dataframe.data[0] is H'32,'2'
rx_dataframe.data[0] is H'33,'3'
I need to to convert hex values 0x00,0x31,0x32,0x33
in to char value '0','0','3','1','3','2';'3','3'
and to store them at the locations of tx_buff_data[];
I want my tx_buff_data
look like this
tx_buff_data[0] have H'30,'0'
tx_buff_data[1] have H'30,'0'
tx_buff_data[2] have H'33,'3'
tx_buff_data[3] have H'31,'1'
tx_buff_data[4] have H'33,'3'
tx_buff_data[5] have H'32,'2'
tx_buff_data[6] have H'33,'3'
tx_buff_data[7] have H'33,'3'
You can split each byte into two nibbles (4-bit quantities) using bitwise AND + shifts:
unsigned char lo = byte & 0x0f;
unsigned char hi = (byte >> 4) & 0x0f;
Then, you can convert each half into a hex character by an array lookup (since strings are just character arrays):
char loChar = "0123456789ABCDEF"[lo];
char hiChar = "0123456789ABCDEF"[hi];
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