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IP address in a variable

kristijanekej (1)
I'm new to C++

I want to know how i can use cin command to input a ip in a variable, i tried it a few time but didn't manage to do it, i made a little research and didn't found much, i found that i need o make the ip in a array but i don't know how to do it, i want it to be in a array bi only entering the whole ip in one cin if that's possible.
Thank you

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#include <iostream>
using namespace std;

int main() {
	int ip;
	cout << "IP adress: ";
	cin >> ip;
	return 0;
 }
jlb (4973)
I suggest you start by getting the address as a string then validating and converting this string to an actual value. And you probably shouldn't be using a single int to try to hold this value.
MiiNiPaa (8886)
How would you want to input IP? A single number? Four octets? How do you want to store it?

Assuming you want to enter octets:
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#include <iostream>
#include <cstdint>

int main()
{
    int octets[4];
    std::uint32_t ip = 0;
    std::cout << "IP adress: ";
    for(int i = 0; i < 4; ++i) {
        if(i != 0) std::cin.get();
        std::cin >> octets[i];
        ip <<= 8u;
        ip += octets[i];
    }
    std::cout << "ip: " << ip << "\nOctets: ";
    for(int i = 0; i < 4; ++i) {
        if(i != 0) std::cout << '.';
        std::cout << +octets[i];
    }
}
 
IP adress: 127.0.0.1
ip: 2130706433
Octets: 127.0.0.1


@jib Actually IPv4 address is a single 32bit unsigned number. It is just separated into 4 octets for display.
Last edited on
jlb (4973)
My point was that an int is not guaranteed to be large enough to hold the value of an IP address.
LB (13399)
std::uint32_t is guaranteed to be the correct size to hold an IPv4 address.

I don't think any compilers support a native type that can hold an IPv6 address.
Last edited on
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