C++ 在不重新解释转换的情况下将无符号字符转换为 std::string 的方法？
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Way to get unsigned char into a std::string without reinterpret_cast?
提问by WilliamKF
I have an unsigned char array that I need in a std::string, but my current way uses reinterpret_cast which I would like to avoid. Is there a cleaner way to do this?
我在 std::string 中有一个我需要的无符号字符数组，但我目前的方式使用了我想避免的 reinterpret_cast 。有没有更干净的方法来做到这一点？
unsigned char my_txt[] = {
0x52, 0x5f, 0x73, 0x68, 0x7e, 0x29, 0x33, 0x74, 0x74, 0x73, 0x72, 0x55
}
unsigned int my_txt_len = 12;
std::string my_std_string(reinterpret_cast<const char *>(my_txt), my_txt_len);
回答by Steve Jessop
Use the iterator constructor:
使用迭代器构造函数：
std::string my_std_string(my_txt, my_txt + my_txt_len);
This is assuming that you want the unsigned chars to be converted to char. If you wantthem to be reinterpreted, then you should use reinterpret_cast. That would be perfectly clean, since what you say is exactly what is done.
这是假设您希望将无符号字符转换为字符。如果您希望重新解释它们，那么您应该使用reinterpret_cast. 那将是完全干净的，因为您所说的正是所做的。
In your example, though, it doesn't make any difference, because all of the values in your array are within the range 0to CHAR_MAX. So it's guaranteed that those values are represented the same way in charas they are in unsigned char, and hence that reinterpreting them is the same as converting them. If you had values greater then CHAR_MAXthen implementations are allowed to treat them differently.
但是，在您的示例中，它没有任何区别，因为数组中的所有值都在0to范围内CHAR_MAX。因此可以保证这些值的表示方式char与它们在中的表示方式相同unsigned char，因此重新解释它们与转换它们是相同的。如果您有更大的值，则CHAR_MAX允许实现以不同的方式对待它们。
回答by Robben_Ford_Fan_boy
Have you tried sstream?
你试过 sstream 吗？
stringstream s;
s << my_txt;
string str_my_txt = s.str();
回答by Praveer Kumar
int _tmain(int argc, _TCHAR* argv[])
{
unsigned char temp = 200;
char temp1[4];
sprintf(temp1, "%d", temp);
std::string strtemp(temp1);
std::cout<<strtemp.c_str()<<std::endl;
return 0;
}
