将uint8转换为字符串

Question

我想将uint8转换为string以将其与另一个进行比较。我的功能得到我的Mac Address，现在我想保存在一个字符串。

int main(int argc, char *argv[]) 
{ 
    kern_return_t kernResult = KERN_SUCCESS; 
    io_iterator_t intfIterator; 
    UInt8   MACAddress[kIOEthernetAddressSize]; 


    kernResult = FindEthernetInterfaces(&intfIterator); 

    if (KERN_SUCCESS != kernResult) { 
     // printf("FindEthernetInterfaces returned 0x%08x\n", kernResult); 
    } 
    else { 
     kernResult = GetMACAddress(intfIterator, MACAddress, sizeof(MACAddress)); 

     if (KERN_SUCCESS != kernResult) { 
     // printf("GetMACAddress returned 0x%08x\n", kernResult); 
     } 
     else { 
      printf("This system's built-in MAC address is %02x:%02x:%02x:%02x:%02x:%02x.\n", 
        MACAddress[0], MACAddress[1], MACAddress[2], MACAddress[3], MACAddress[4], MACAddress[5]); 
     } 
    } 

    (void) IOObjectRelease(intfIterator); // Release the iterator. 

    return kernResult; 
} 

我该怎么办？我一直在寻找帮助，但没有任何工作。我在Xcode上。

Answer 1

实际上，您将uint8_t的数组转换为字符串。该规范的方法是使用字符串流：

std::stringstream ss; 
for (size_t i = 0; i < 6; ++i) { 
    ss << MACAddress[i]; 
    if (i != 5) ss << ":"; 
} 
std::string MACstring = ss.str(); 

你可能避免这种通过使用to_string和串联：

std::string MACstring; 
for (size_t i = 0; i < 6; ++i) { 
    MACstring += std::to_string(MACAddress[i]); 
    if (i != 5) MACstring += ":"; 
}