Input
Output
(Sorry for not being detailed)
Answer:
a) Yes
b) Yes
c) Yes
d) No
e) Yes
f) No
Explanation:
a) All single-bit errors are caught by Cyclic Redundancy Check (CRC) and it produces 100 % of error detection.
b) All double-bit errors for any reasonably long message are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
c) 5 isolated bit errors are not caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit since CRC may not be able to catch all even numbers of isolated bit errors so it is not even.
It produces nearly 100 % of error detection.
d) All even numbers of isolated bit errors may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
e) All burst errors with burst lengths less than or equal to 32 are caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit. It also produces 100 % of error detection.
f) A burst error with burst length greater than 32 may not be caught by Cyclic Redundancy Check (CRC) during the transmission of 1024 bit.
Cyclic Redundancy Check (CRC) does not detect the length of error burst which is greater than or equal to r bits.
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! :)
The answer is IPS (Intrusion Prevention Systems)
The Intrusion Prevention Systems and Intrusion Detection Systems (IDS) are two security technologies that secure networks and are very similar in how they work. The IDS detects unauthorized user activities, attacks, and network compromises, and also alerts. The IPS, on the other hand, as mentioned, is very similar to the IDS, except that in addition to detecting and alerting, it can also takes action to prevent breaches.
Answer: bar code
Explanation: that is what it is
