Answer:
No. This is not more secure.
Explanation:
this is not more secure than having all of Carol, bob and Alice having the same key.
alice would be able to know Carol's key for her to be able to verify the answer Carol gave to a challenge by her. she would also have to know bob's key to do the same.
any of these 3 would have to know someone else's secret key to verify answers to any challenge.
the challenge is to know the secret keys if the other two and if done, decryption can easily be done and they can impersonate.
Answer:
see explaination
Explanation:
#include<stdio.h>
/* Your solution goes here */
//Impllementation of SwapArrayEnds method
void SwapArrayEnds(int sortArray[],int SORT_ARR_SIZE){
//Declare tempVariable as integer type
int tempVariable;
if(SORT_ARR_SIZE > 1){
tempVariable = sortArray[0];
sortArray[0] = sortArray[SORT_ARR_SIZE-1];
sortArray[SORT_ARR_SIZE-1] = tempVariable;
}
}
int main(void) {
const int SORT_ARR_SIZE = 4;
int sortArray[SORT_ARR_SIZE];
int i = 0;
sortArray[0] = 10;
sortArray[1] = 20;
sortArray[2] = 30;
sortArray[3] = 40;
SwapArrayEnds(sortArray, SORT_ARR_SIZE);
for (i = 0; i < SORT_ARR_SIZE; ++i) {
printf("%d ", sortArray[i]);
}
printf("\n");
return 0;
}
Please go to attachment for the program screenshot and output
Answer:
answer is D
Explanation:
as the block returns the number of array members, the most likely scenario is the last one.