<span>Not a valid IPv6 address
A valid IPv6 address consist of 8 groups of 4 hexadecimal numbers separated by colons ":". But that can make for a rather long address of 39 characters. So you're allowed to abbreviate an IPv6 address by getting rid of superfluous zeros. The superfluous zeros are leading zeros in each group of 4 digits, but you have to leave at least one digit in each group. The final elimination of 1 or more groups of all zeros is to use a double colon "::" to replace one or more groups of all zeros. But you can only do that once. Otherwise, it results in an ambiguous IP address. For the example of 2001:1d5::30a::1, there are two such omissions, meaning that the address can be any of
2001:1d5:0:30a:0:0:0:1
2001:1d5:0:0:30a:0:0:1
2001:1d5:0:0:0:30a:0:1
And since you can't determine which it is, it's not a valid IP address.</span>
B. Television. It provided <span>instant communication and information to a massive audience for the first time in 1927.
Hope this helps :)</span>
Answer:
#include <iostream>
using namespace std;
void divide(int numerator, int denominator, int *quotient, int *remainder)
{
*quotient = (int)(numerator / denominator);
*remainder = numerator % denominator;
}
int main()
{
int num = 42, den = 5, quotient=0, remainder=0;
divide(num, den, "ient, &remainder);
return 0;
}
Explanation:
The exercise is for "Call by pointers". This technique is particularly useful when a variable needs to be changed by a function. In our case, the quotient and the remainder. The '&' is passing by address. Since the function is calling a pointer. We need to pass an address. This way, the function will alter the value at the address.
To sum up, in case we hadn't used pointers here, the quotient and remainder that we set to '0' would have remained zero because the function would've made copies of them, altered the copies and then DELETED the copies. When we pass by pointer, the computer goes inside the memory and changes it at the address. No new copies are made. And the value of the variable is updated.
Thanks! :)
Answer:
To make the group of four we can write 100011 as 00100011
Now, the two groups are:- 0010 and 0011
And 0010 in binary corresponds to 3 in Hexadecimal
And 0011 in binary corresponds to 4 in Hexadecimal.
So, 100011 of binary corresponds to 34 of hex.
False, forsure mates but gotta love the question:)
