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MAXImum [283]
3 years ago
12

public class Test { public static void main(String[] args) { try { method(); System.out.println("After the method call"); } catc

h (ArithmeticException ex) { System.out.println("ArithmeticException"); } catch (RuntimeException ex) { System.out.println("RuntimeException"); } catch (Exception e) { System.out.println("Exception"); } } static void method() throws Exception { System.out.println(1 / 0); } }
Computers and Technology
1 answer:
USPshnik [31]3 years ago
8 0

Answer:

"ArithmeticException" is the correct answer for the above question.

Explanation:

Missing Information : The above question does not hold that "what is the output of the program".

  • The above question has a class that holds the two functions one function is the main function which calls the other function in a try-catch block.
  • The method function holds one run time exception which is 1/0 which is an arithmetic exception.
  • It can be handle by the help of an arithmetic class object which is defined in the first catch block.
  • Hence the print function of this catch is executed and prints "ArithmeticException".
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A simple operating system supports only a single directory but allows it to have arbitrarily many files with arbitrarily long fi
Elden [556K]

Answer:

Yes

Explanation:

Yes, something approximating a hierarchical file system be simulated. This is done by assigning to each file name the name of the directory it is located in.

For example if the directory is UserStudentsLindaPublic, the name of the file can be UserStudentsLindaPublicFileY.

Also the file name can be assigned to look like the file path in the hierarchical file system. Example is /user/document/filename

7 0
3 years ago
Write a loop that sets newScores to oldScores shifted once left, with element 0 copied to the end. Ex: If oldScores = {10, 20, 3
Katarina [22]

Answer:

The code is as follows:

for(int j = 0; j < newScores.length-1; j++){  

               newScores[j] = oldScores[j+1];  

           }  

           newScores[oldScores.length-1] = oldScores[0];

Explanation:

This loop iterates through the elements of oldScores

for(int j = 0; j < newScores.length-1; j++){  

This enters the elements of oldScores to newScores starting from the element at index 1

               newScores[j] = oldScores[j+1];  

           }  

This moves the first element of index 0 to the last index of newScores

           newScores[oldScores.length-1] = oldScores[0];

7 0
3 years ago
If Nancy receives an encrypted message from Matthew, which key does she use to read it? Nancy’s private key Nancy’s public key M
Juliette [100K]

Answer:

Nancy's private key

Explanation:

She's receiving it.  That means Matthew used her public key to encrypt it.  She should decrypt it using her private key.  Watch this video for clarification :)

Watch The Internet: Encryption & Public Keys on the big video site which cannot be named on brainly :p

7 0
4 years ago
Determine whether or not the following pairs of predicates are unifiable. If they are, give the most-general unifier and show th
Evgen [1.6K]

Answer:

a) P(B,A,B), P(x,y,z)

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }.

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q.

c. Older(Father(y),y), Older(Father(x),John)

Thus , mgu ={ y/x , x/John }.

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B.

e) P(f(x), x, g(x)), P(f(y), A, z)    

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

Explanation:  

Unification: Any substitution that makes two expressions equal is called a unifier.  

a) P(B,A,B), P(x,y,z)  

Use { x/B}  

=> P(B,A,B) , P(B,y,z)  

Now use {y/A}  

=> P(B,A,B) , P(B,A,z)  

Now, use {z/B}  

=> P(B,A,B) , P(B,A,B}  

Hence, most general unifier = {x/B , y/A , z/B }  

b. P(x,x), Q(A,A)  

No mgu exists for this expression as any substitution will not make P(x,x), Q(A, A) equal as one function is of P and the other is of Q  

c. Older(Father(y),y), Older(Father(x),John)  

Use {y/x}  

=> Older(Father(x),x), Older(Father(x),John)  

Now use { x/John }  

=> Older(Father(John), John), Older(Father(John), John)  

Thus , mgu ={ y/x , x/John }  

d) Q(G(y,z),G(z,y)), Q(G(x,x),G(A,B))  

Use { y/x }  

=> Q(G(x,z),G(z,x)), Q(G(x,x),G(A,B))

Use {z/x}  

=> Q(G(x,x),G(x,x)), Q(G(x,x),G(A,B))  

This is not unifiable as x cannot be bound for both A and B  

e) P(f(x), x, g(x)), P(f(y), A, z)  

Use {x/y}  

=> P(f(y), y, g(y)), P(f(y), A, z)  

Now use {z/g(y)}  

P(f(y), y, g(y)), P(f(y), A, g(y))  

Now use {y/A}  

=> P(f(A), A, g(A)), P(f(A), A, g(A))  

Thus , mgu = {x/y, z/y , y/A }.

7 0
3 years ago
The excessive use of the internet and the web for personal use at work is sometimes called ____ .
Elodia [21]
This is called Cyberslacking
7 0
3 years ago
Read 2 more answers
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