十六进制转换问题，即刻送分

[z_yongqiang]

如何将字符串"C2"转换为十六进制数0xC2

[fatwave]

int communication::HtoD(unsigned char *hex_str, unsigned char *bin_str)
/*
"ba087d4012345678" -> {0xba, 0x08, 0x7d, 0x40, 0x12, 0x34, 0x56, 0x78}
For ASCII string and OS, not for EBCDIC string and OS.
Return: > 0 --- bin_str length
< 0 --- error

M.L.Y 1999.12.28
*/
{
int i, j, n;
int c1, h1, h2;
unsigned char *sh = hex_str, *sb = bin_str;

n = strlen(sh);
if(n < 2) return -1;
if(n % 2 != 0) return -2;
for(i = 0, j = 0; i < n; j++)
{
if((h1 = hex_to_num(sh[i++])) < 0) return -1;
if((h2 = hex_to_num(sh[i++])) < 0) return -1;
c1 = h1 << 4 | h2;
sb[j] = c1;
}
return j;
}

[Ozzman]

int atoi( const char *string );

看MSDN帮助！给分！

[fatwave]

int communication::hex_to_num(char hex1char)
/*
'a' or 'A' -> 10, '1' -> 1
If hex1char not in '0'-'9','A'-'F','a'-'f' return -1
*/
{
if(hex1char >= '0' && hex1char <= '9') return hex1char - '0';
if(hex1char >= 'A' && hex1char <= 'F') return hex1char - 'A' + 10;
if(hex1char >= 'a' && hex1char <= 'f') return hex1char - 'a' + 10;
return -1;
}

[bcpl]

int iVal;
char *szVal = "C2";
sscanf(szVal, "%x", &iVal);

[bcpl]

int iVal;
char *szVal = "C2";
sscanf(szVal, "%x", &iVal);

[wuxfBrave]

char szDigit[3] = { 'C', '2' };
int nLen = _tcslen(szDigit);
int nOut = 0;
for (int i = nLen-1; i >= 0; i--)
{
if (szDigit[i] >= 'A' && szDigit[i] <= 'F')
{
nOut += (szDigit[i] -'A'+10)*pow(16, nLen-1-i);
}
else
{
nOut += (szDigit[i]-'0')*pow(16, nLen-1-i);
}
}