objective-c - 如何将具有十六进制值的 NSString 转换为二进制 char*？

Question

我有一个很长的十六进制值存储为 NSString，如下所示:

c12761787e93534f6c443be73be31312cbe343816c062a278f3818cb8363c701

如何将其转换回存储为 char* 的二进制值

Answer 1

这有点草率，但应该能让你走上正轨:

NSString *hexString = @"c12761787e93534f6c443be73be31312cbe343816c062a278f3818cb8363c701";
char binChars[128]; // I'm being sloppy and just making some room
const char *hexChars = [hexString UTF8String];
NSUInteger hexLen = strlen(hexChars);
const char *nextHex = hexChars;
char *nextChar = binChars;
for (NSUInteger i = 0; i < hexLen - 1; i++)
{
    sscanf(nextHex, "%2x", nextChar);
    nextHex += 2;
    nextChar++;
}