一秒钟问题急救，将字符转换成十六进制与将16进制转换为字符的函数？

[MiracleRecurrence]

100分急救，将字符转换成16进制与将16进制转换为字符的函数？

[sxbSun]

DWORD dwPid;
dwPid=0;
//CString svPid = "FFF9AA1D";
dwPid = _tcstoul((LPSTR)(LPCTSTR)svPid, NULL,16); //16进制
上面为将字符转换成十六进制。

下面位将16进制转换为字符
wsprintf(svBuffer,"PROC: (0x%X) %s %d threads",cur->dwProcID,cur->svApp,nThreads);

给分吧！！！??

[gameboy999]

倒...原来有那么一个函数,我还自己写了一个...晕菜
int a16toi(char * pNumber)
{
int ret = 0;
int len = strlen(pNumber);
int pos = len;
while(pos--)
switch(pNumber[pos])
{
case '0': break;
case '1': ret += (len-pos-1) ? 1 * 1<<(4* (len-pos-1)) : 1;break;
case '2': ret += (len-pos-1) ? 2 * 1<<(4* (len-pos-1)) : 2;break;
case '3': ret += (len-pos-1) ? 3 * 1<<(4* (len-pos-1)) : 3;break;
case '4': ret += (len-pos-1) ? 4 * 1<<(4* (len-pos-1)) : 4;break;
case '5': ret += (len-pos-1) ? 5 * 1<<(4* (len-pos-1)) : 5;break;
case '6': ret += (len-pos-1) ? 6 * 1<<(4* (len-pos-1)) : 6;break;
case '7': ret += (len-pos-1) ? 7 * 1<<(4* (len-pos-1)) : 7;break;
case '8': ret += (len-pos-1) ? 8 * 1<<(4* (len-pos-1)) : 8;break;
case '9': ret += (len-pos-1) ? 9 * 1<<(4* (len-pos-1)) : 9;break;
case 'a': ret += (len-pos-1) ? 10 * 1<<(4* (len-pos-1)) : 10;break;
case 'b': ret += (len-pos-1) ? 11 * 1<<(4* (len-pos-1)) : 11;break;
case 'c': ret += (len-pos-1) ? 12 * 1<<(4* (len-pos-1)) : 12;break;
case 'd': ret += (len-pos-1) ? 13 * 1<<(4* (len-pos-1)) : 13;break;
case 'e': ret += (len-pos-1) ? 14 * 1<<(4* (len-pos-1)) : 14;break;
case 'f': ret += (len-pos-1) ? 15 * 1<<(4* (len-pos-1)) : 15;break;
case 'A': ret += (len-pos-1) ? 10 * 1<<(4* (len-pos-1)) : 10;break;
case 'B': ret += (len-pos-1) ? 11 * 1<<(4* (len-pos-1)) : 11;break;
case 'C': ret += (len-pos-1) ? 12 * 1<<(4* (len-pos-1)) : 12;break;
case 'D': ret += (len-pos-1) ? 13 * 1<<(4* (len-pos-1)) : 13;break;
case 'E': ret += (len-pos-1) ? 14 * 1<<(4* (len-pos-1)) : 14;break;
case 'F': ret += (len-pos-1) ? 15 * 1<<(4* (len-pos-1)) : 15;break;
default:
break;
}
return ret;
}

[MiracleRecurrence]

谢了！！！

[aileen_long]

_itoa(int value, char *buffer, int radix);
// 把整型转换成字符串，radix=10，转换为十进制格式，radix=16，转换为十六进制格式
int atoi(const char *string);
// 把字符串传进去，返回int
long atol(const char *string);
// 把字符串传进去，返回long
double atof(const char *string);
// 把字符串传进去，返回double
更多描述参考MSDN

[aileen_long]

你给分还真是时候！俺正在写的时候，你给分了！

[MiracleRecurrence]

aileen_long(期待2002) ,别伤心，以后有大把机会，我很大方的