c++ - unsigned char* 值转为十六进制

Question

这个问题已经有答案了:

Convert two ASCII Hexadecimal Characters (Two ASCII bytes) in one byte (6 个回答)

已关闭 7 年前。

我有一个典型的用例，我需要将无符号字符值转换为十六进制。

例如，

unsigned char *pBuffer = (unsigned char *)pvaluefromClient //where pvaluefromclient is received from a client

pBuffer的长度是32字节，它保存的值如下，

(gdb) p pBuffer 
$5 = (unsigned char *) 0x7fd4b82cead0 "EBA5F7304554DCC3702E06182AB1D487"
(gdb) n

第 1 步:我需要按如下方式拆分此 pBuffer 值，

{EB,A5,F7,30,45,54,DC,C3,70,2E,06,18,2A,B1,D4,87 }

第2步:我需要将上面的分割值转换为十进制，如下所示，

const unsigned char pConvertedBuffer[16] = {
    235,165,247,48,69,84,220,195,112,46,6,24,42,177,212,135
};

关于如何实现 STEP1 和 STEP2 有什么想法吗？对此的任何帮助将不胜感激

Answer 1

像这样怎么样:

unsigned char *pBuffer = (unsigned char *)pvaluefromClient //where valuefromclient is received from a client

int i, j;
unsigned char target[16]
for(i=0;i<32;i+=2)
{
   sscanf((char*)&pBuffer[i], "%02X", &j);
   target[i/2] = j;
}