The question is incomplete as the scenarios are not given that were to be matched by the given encryption algorithms.
By finding the question from internet i have attached the image of the scenarios in the form of table and answered below accordingly
<h2>
Answer:</h2>
<u>Scenario A:</u>
Advanced Encryption Standard (AES) is a technique that uses 256-bit cipher texts for encryption process. The data encrypted by AES cannot be attacked and even it cannot be decrypted unless a brute-force search is used through all of possible 256-bit keys.For mobile security it is known as one of the best algorithms.
<u>Scenario B:</u>
Following are the techniques of encryption used for given scenario:
- ECC
- Digital Signature
- AES
- Blowfish
<u>Scenario C:</u>
Following are the techniques of encryption used for given scenario:
<u>Scenario D:</u>
Secure Sockets Layer (SSL) - SSL is a protocol that is commonly-used for managment of the message transmission security over the Internet)
<u>Scenario E:</u>
Advanced Encryption Standard (AES) - AES is an encryption algorithm used by U.S. Government agencies for securing sensitive but unclassified material .
i hope it will help you!
Answer:
in my opinion 4
Explanation:
when the system is available to users
(sorry and thanks)
Answer:
I get 0x55 and this the linking address of the main function.
use this function to see changes:
/* bar6.c */
#include <stdio.h>
char main1;
void p2()
{
printf("0x%X\n", main1);
}
Output is probably 0x0
you can use your original bar6.c with updaated foo.c
char main;
int main() // error because main is already declared
{
p2();
//printf("Main address is 0x%x\n",main);
return 0;
}
Will give u an error
again
int main()
{
char ch = main;
p2(); //some value
printf("Main address is 0x%x\n",main); //some 8 digit number not what printed in p2()
printf("Char value is 0x%x\n",ch); //last two digit of previous line output
return 0;
}
So the pain in P2() gets the linking address of the main function and it is different from address of the function main.
Now char main (uninitialized) in another compilation unit fools the compiler by memory-mapping a function pointer on a char directly, without any conversion: that's undefined behavior. Try char main=12; you'll get a multiply defined symbol main...
Explanation:
Answer:validity
Explanation:
Because it dont sound right