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Ganezh [65]
2 years ago
5

Explain the distinction between a real address and a virtual address.

Computers and Technology
2 answers:
NARA [144]2 years ago
8 0

Answer:

send to you and the world is

kap26 [50]2 years ago
6 0

Answer:

<em />

<em>A real address is the physical address in main memory. Though a virtual address is the address of a storage location in virtual memory.</em>

<em></em>

<em></em>

<em>hopes this help (:`</em>

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7. Write a program in C to display the string "ARRAY" in the following format : A AR ARR ARRA ARRAY
Art [367]
This program will the string array in the given format.

int main(void){
char arr[6];
int counter;
strcpy(arr[], "ARRAY");

for(counter=0; counter<6; counter++){
printf("%c", arr[counter]);
}

return 0;
)

Note: Do not forget to include all the necessary library that is needed to run this program.

6 0
3 years ago
Hi I paid for brainy plus, but it says the it can't renew and try paying for it again but it won't work. Please help
Scorpion4ik [409]

Answer:

Contact CS, ( customer support ) They should be able to help you. Go to this link for more info. brainly.com/contact/index

Explanation:

8 0
4 years ago
he function below takes one parameter: a list of strings (string_list). Complete the function to return a new list containing on
Aleks04 [339]

Answer:

def select_short_strings(string_list):

   new_list = []

   

   for s in string_list:

       if len(s) < 20:

           new_list.append(s)

   return new_list

   

lst = ["apple", "I am learning Python and it is fun!", "I love programming, it is easy", "orange"]

print(select_short_strings(lst))

Explanation:

- Create a function called <em>select_short_strings</em> that takes one argument <em>string_list</em>

Inside the function:

- Initialize an empty list to hold the strings that are less than 20

- Inside the loop, check the strings inside <em>string_list</em> has a length that is smaller than 20. If found one, put it to the <em>new_list</em>.

- When the loop is done, return the <em>new_list</em>

- Create a list to check and call the function

4 0
4 years ago
Read 2 more answers
Process redesign and reengineering in 1500 words
timama [110]

Answer:

Process re-design is the process in which re-work is done to achieve the improvement in the output. The major aim is to set the goals for achieving the main objective.

The various essential steps in the process of re-design are:

  • Data capturing process are used to identify the process and making the necessary changes.
  • Identify every step of the process to ensure that process is in organized way along with the workflow.

Re-engineering is the process of re-design the system with focuses to improve the efficiency and quality. It basically analysis the whole business re-engineering process. The main step is to identify the need for the changes.

 

7 0
3 years ago
Using MARS/MIPS
slega [8]

Answer:

Explanation:

MIPS program which increments from 0 to 15 and display results in Decimal on the console

In this program the user defined procedures print_int and print_eot were used to print the integer values and new line characters(\n) respectively.the mechanisam of the loop is explaine in the comment section of the program.

     addi $s0, $0, 0

     addi $s1, $0, 15

print_int:

              li $v0, 1                # system call to print integer

              syscall                

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s1<$s0

beq $s0, $0, end

end:

MIPS progam to increment from 0 to 15 and display results in Hexadecimal on the console

this program is slightly differed from the previous program in this program the system call issued in print_int is implemented with a system call that prints numbers in hex.

addi $s0, $0, 15

     addi $s1, $0, 0

print_int:

   li      $v0,34                  # syscall number for "print hex"

   syscall                         # issue the syscall

              jr $ra                     # return

print_eol:                      # prints "\n"

            li $v0, 4            

              la $a0, linebrk      

              syscall              

              jr $ra                  # return

main: . . .          

              li $a0, 0                # print 0

              jal print_int         # print value in $a0

loop:   move $a0, $s0           # print loop count

       jal print_int        

       jal print_eol           # print "\n" character

       addi $s0, $s0, 1        # increment loop count by 1

       ble $s1, $s0, loop      # exit if $s0>$s1

beq $s0, $0, end

end:

5 0
3 years ago
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