十六进制字符串转char数组
RT、例如我输入"1A023b4d5c6f9",没有任何分符，如何如何获取这个十六进制的字符串，然后转换成unsigned char的数组。

例如：
unsigned char tmp[7];
tmp[0]=0x1A;
tmp[1]=0x02;
.........这样
[解决办法]
先每两个字符截取出来，再转换成数值呗
[解决办法]

        CString str(_T("1A023b4d5c6f9"));
        char tmp[6];
char arr[2];
int j = 0;
int k = 0;
for(int i=0; i<str.GetLength(); i++)
{
if (j == 0)
{
arr[j] = str.GetAt(i);
j = 1;
}
else
{
arr[j] = str.GetAt(i);
int tempNum = atoi(arr);
tmp[k] = tempNum/10*16 + tempNum%10;
k++;
j = 0;
}
}

[解决办法]

    CString str(_T("1A023b4d5c6f9"));
    char tmp[7];
    char arr[2];
    int j = 0;
    int k = 0;
    for(int i=0; i<str.GetLength(); i++)
    {
        if (j == 0)
        {
            arr[j] = str.GetAt(i);
            j = 1;
        }
        else
        {
            arr[j] = str.GetAt(i);
            int tempNum = atoi(arr);
            tmp[k] = tempNum/10*16 + tempNum%10;
            k++;
            j = 0;
        }
    }
tmp[6] = '\0';

[解决办法]
CString str(_T("1A023b4d5c6f9"));
str.GetBufferSetLength(2);
int nPos = _wtoi(str);
char temp = nPos;

以上只是例子
[解决办法]

CString str(_T("1A023b4d5c6f9"));
CString sTemp = str;
int nlen = str.GetLength()/2;\\考虑这个能否整除
char *Buf = new char[nlen];
ZeroMemory(Buf,nlen);
for (int i =0;i<nlen;i++)
{
sTemp.GetBufferSetLength(2);
Buf =(char) _wtoi(sTemp);
str.Delete(0,2);
sTemp.ReleaseBuffer();

}

[解决办法]

引用:

先每两个字符截取出来，再转换成数值呗

美女 求交往 教我编程！
[解决办法]
全部读进来，然后两两截取，再转换保存到char里面。
[解决办法]

short int HexChar2Dec(char c)  
{  
    if ( '0'<=c && c<='9' ) {  
        return short(c-'0');  
    } else if ( 'a'<=c && c<='f' ) {  
        return ( short(c-'a') + 10 );  
    } else if ( 'A'<=c && c<='F' ) {  
        return ( short(c-'A') + 10 );  
    } else {  
        return -1;  
    }  
} 

int str2Num16(const char* str)
{
  return (str[i+1] == '\0') ? HexChar2Dec(str[i]) : HexChar2Dec(str[i])*16+HexChar2Dec(str[i+1]);
}
void Parse()
{
         char hexstr[] = "1A023b4d5c6f9";
int len = (strlen(hexstr)%2 == 0) ? strlen(hexstr)/2 : (strlen(hexstr)/2+1);
unsigned char hex[128] = {0};
for (int i = 0; i < len; ++i)
{
int idx = i*2;
hex[i] = str2Num16(hexstr+idx);
}
}

------解决方案--------------------

引用:

short int HexChar2Dec(char c)  
{  
    if ( '0'<=c && c<='9' ) {  
        return short(c-'0');  
    } else if ( 'a'<=c && c<='f' ) {  
        return ( short(c-'a') + 10 );  
    } else if ( 'A'<=c && c<='F' ) {  
        return ( short(c-'A') + 10 );  
    } else {  
        return -1;  
    }  
} 

int str2Num16(const char* str)
{
  return (str[i+1] == '\0') ? HexChar2Dec(str[i]) : HexChar2Dec(str[i])*16+HexChar2Dec(str[i+1]);
}
void Parse()
{
         char hexstr[] = "1A023b4d5c6f9";
int len = (strlen(hexstr)%2 == 0) ? strlen(hexstr)/2 : (strlen(hexstr)/2+1);
unsigned char hex[128] = {0};
for (int i = 0; i < len; ++i)
{
int idx = i*2;
hex[i] = str2Num16(hexstr+idx);
}
}

更正一下
int str2Num16(const char* str)
{
return (str[1] == '\0') ? HexChar2Dec(str[0]) : HexChar2Dec(str[0])*16+HexChar2Dec(str[1]);
}