I have a program which im writing that will count up ip addresses starting from 1.0.0.1. Each octet of the ip address is assigned as an int. ie int oct1, int oct2 I then want to assign these 4 values to make one address which is assigned as a char. My program keeps telling my cannot convert int to char. Any ideas? Cheers

To add: I know you can use sprintf to print it, but im not actually wanting to print it.

Char is 8 bits. Int is 16 bits. Therefore you cannot get 16 bits into 8 bits without data overflow. So why not use Char to begin with. A char can have maximum value of 255 which is all you need for an IP octet. The full IPA address is 32 bits which needs a Long Int. You can put 4 char data types into a Long Int. Alternatively just use a Long Int and use bit masking to isolate the four octets. Stuart

try this out...sprintf doesn't print to an output device, it alters a variable. This made my laptop screen get all frizzy when it was running...it's not my fault if you blow something up. I made cout print to the screen so you could see what was going in...replace that with what you want to do with the ipAddress variable. #include <iostream> using namespace std; int main() { char ipAddress[16] = " "; int oct1 = 1; int oct2 = 0; int oct3 = 0; int oct4 = 1; while (oct1 <= 255) { while (oct2 <= 255) { while (oct3 <= 255) { while (oct4 <= 255) { sprintf(ipAddress, "%d.%d.%d.%d", oct1, oct2, oct3, oct4); cout << endl << ipAddress; oct4++; } oct3++; oct4 = 0; } oct2++; oct3 = 0; } oct1++; oct2 = 0; } cin.get(); cin.get(); return 0; }