将十六进制转换为64位bitset c ++

Question

I want to input a string of up to 16 hex characters to the screen and then convert the string to a bitset<64>

So far I have managed the following

string tempString;
unsigned int tempValue;

cout << "enter addr in hex : ";
cin >> tempString;
istringstream ost(tempString);
ost >> hex >> tempValue;
bitset<32>  addr(tempValue);
cout << "addr = " << addr << endl;

which works fine, but then when I repeat for 64 bit it fails. Playing around it seems to only fail for the top 32 bits!

bitset<64> wdata = 0;
if (rdnwr[0] == 0)
{
    cout << "enter wdata in hex : ";
    cin >> tempString;
    istringstream ost1(tempString);
    ost1 >> hex >> tempValue;
    wdata = tempValue;
    cout << "wdata = " << wdata << endl;
}

Is this to do with a maximum size for the istringstream? Or perhaps the different way I am assigning wdata?

Thanks.

Answer 1

At a guess, you missed changing something to 64 bits (either the bitset, or possibly changing the int to long long ). This, however:

string tempString;
unsigned long long tempValue;

cout << "enter addr in hex : ";
cin >> tempString;
istringstream ost(tempString);
ost >> hex >> tempValue;
bitset<64>  addr(tempValue);
cout << "addr = " << addr << endl;

...seems to work, at least for me:

enter addr in hex : 0123456789abcdef
addr = 0000000100100011010001010110011110001001101010111100110111101111

[tested with both VC++ and MinGW, with identical results]