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Alexxx [7]
3 years ago
5

Why it is so important for all application builders to always check data received from unknown sources, such as web applications

, before using that data?
Computers and Technology
1 answer:
Aleks04 [339]3 years ago
6 0
The already-long problem on hacking and other security-related cases necessitates for the application builders to check the data received from unknown sources. To further strengthen the security, even data received from known sources also need to be check as these may also contain harmful viruses creating critical problem to the application. 
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By apply styles, _____________ formats are being applied each time.
Gnoma [55]
B. different is the correct answer.
5 0
3 years ago
The university computer lab’s director keeps track of lab usage, as measured by the number of students using the lab. This funct
Brut [27]

Answer:

to keep count of how many users there are in total.

Explanation:

all i had to do was read the question twice to understand the answer is pretty

much in the question.

4 0
4 years ago
_______________ is used by a hacker to mask intrusion and obtain administrator permissions to a computer.
Sergeu [11.5K]

Answer:

Rootkit.

Explanation:

Rootkit is a collection of software tools,mostly malicious.These are mostly used by hackers to obtain administrator permission to a computer by masking intrusion.

Root-kit is made from two word Root and kit.Where root refers to the name of privileged account on an operating system that is somewhat like unix and KIT refers to the tools used.

6 0
4 years ago
Let f be the following function: int f(char *s, char *t){char *p1, *p2;for(p1 = s, p2 = t; *p1 != ‘\0’&& *p2 != ‘\0’; p1
siniylev [52]

Answer:

return value =2.

Here the function f() returns the length of the substring we traversed before we find the same character at the equal index of two substrings.

Take the inputs s= “abcd” and t= “bccd”.  

• Now, p1 points to s1, i.e., p1 points to the character ‘a’ of “abcd”. And similarly, p2 points to ‘b’ of “bccd”.

• Then we compare the values at p1 and p2, are not equal, so p1 and p2 both are incremented by 1.

• Now the characters ‘b’ and ‘c’ of “abcd” and “bccd” respectively are compared. They are not equal. So both p1 and p2 both are incremented by 1.

• Now, p1 points to ‘c’ of “ abcd” that is the element at index 2 of s. And p2 points to ‘c’ of “bccd” that is the element at index 2 of t. Here value at p1 and p2 becomes equal. So the break statement is executed. We stop moving forward.

• As p1 is pointing to index 2 and s is pointing to the base that is index 0, so p1-s = 2.

Explanation:

#include<stdio.h>

int f(char *s, char *t);

void main()

{

int k = f("abcd", "bccd");

printf("%d", k);

}

int f(char *s, char *t)

{

char *p1, *p2;

for(p1 = s, p2 = t; *p1 != '\0'&& *p2 != '\0'; p1++, p2++)

{

 if (*p1 ==*p2)  

  break;  

}

return (p1-s);

}

OUPUT is given as image

7 0
3 years ago
Question 19
MaRussiya [10]

Answer

D

Explanation:

5 0
4 years ago
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