The answer is Salted Password Hashing. The process is similar to hashing., but with a twist. A random value is introduced for each user. This salt value<span> is included with the password when the hash value is calculated and is stored with the user record. Including the salt value means that two users with the same password will have different password hashes.</span>
Answer:
See explanation
Explanation:
Replace the ____ with the expressions in bold and italics
1) <em> return km</em>
return km returns the result of the computation
2) = <em>convert_distance(my_trip_miles)</em>
convert_distance(my_trip_miles) calls the function and passes my_trip_miles to the function
3) + <em>str(my_trip_km)</em>
The above statement prints the returned value
4) +str(my_trip_km * 2)
The above statement prints the returned value multiplied by 2
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
This type of security failure is called WIN32K_SECURITY_FAILURE. <span> This indicates a </span>security failure was detected in win32k. <span>According to a new report issued by Dark Reading, there are a number of key </span>security failures<span> that cybercriminals take advantage of.</span>
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:
