Answer:
Digital Signature
Explanation:
Digital Signatures are used in electronic messages to verify the sender's ıdentity. It is a online signature and highly secure way of proving identity.
The Signature is <em>encrypted</em> and can be decoded using <em>public-key</em> .
Digital signatures are certificated uniquely from accredited providers, encrypted and can be validated by certificate authorities.
Messages with digital signatures prove that the message sent by the owner of the signature and didn't changed on the way.
Answer:
The correct answer to the following question is an option (b) parity.
Explanation:
Because it is a bit that appends to the binary bit and it has the sets of one or either zero by which a total number of the one bits is the odd or the even parity bit.
Parity bit is a type of bit by which we can check when the data is moved to another place than at that time the data is written over or lost and also we check data at that time when these data is transmitted from one computer to other computers.
For the first one its NOW, I'm not sure about the 2nd, the 3rd I think is penal code cases but i could be wrong, and the 4th one sounds like voting is the most logical answer.
Answer:
hope this helps. I am also a learner like you. Please cross check my explanation.
Explanation:
#include
#include
using namespace std;
int main()
{
int a[ ] = {0, 0, 0}; //array declared initializing a0=0, a1=0, a3=0
int* p = &a[1]; //pointer p is initialized it will be holding the address of a1 which means when p will be called it will point to whatever is present at the address a1, right now it hold 0.
int* q = &a[0]; //pointer q is initialized it will be holding the address of a0 which means when q will be called it will point to whatever is present at the address a0, right now it hold 0.
q=p; // now q is also pointing towards what p is pointing both holds the same address that is &a[1]
*q=1
; //&a[0] gets overwritten and now pointer q has integer 1......i am not sure abut this one
p = a; //p is now holding address of complete array a
*p=1; // a gets overwritten and now pointer q has integer 1......i am not sure abut this one
int*& r = p; //not sure
int** s = &q; s is a double pointer means it has more capacity of storage than single pointer and is now holding address of q
r = *s + 1; //not sure
s= &r; //explained above
**s = 1; //explained above
return 0;
}
