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2 years ago

What is one way object-oriented programming differs from procedural programming?

1 answer:

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The way object-oriented programming differs from procedural programming is that In object-oriented programming, the programmer thinks in terms of an instance of a class that has its own attributes and behaviors.

<h3>How does object oriented programming differs from procedural programming?</h3>

Object-oriented Programming is known to be one that uses classes and objects, while Procedural Programming is one that takes on applications by handling problems using a top-down approach where it solves problem from the top of the code and to the down to the bottom.

Procedural Programming is known as a programming design approach that tends to divides well defined tasks into modules, functions, etc.

Learn more about Procedural Programming from

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The spreadsheet below shows how much money each store raised for charity during the months of January, February, and March. A1:

Elenna [48]

Answer:

B2:B4

Explanation:

6 0

2 years ago

Read 2 more answers

Since the 1960s, there have been five major generations, or cycles, of computing. Which represents those generations in order?

sergiy2304 [10]

Answer:a) Mainframes, miniframes, PCs, desktop Internet computing, mobile computing

Explanation: The hierarchy of the computers or computing are in the generation is:-

Mainframe:-designed for the execution of huge data processing , storage and execution.
Miniframe:-has similar functioning as the mainframe but on smaller scale and version.
Personal computer(PC):-It is designed for the individual purpose and according the need of the user.
Mobile computing:-computing that can be done on the mobile phone similar to the computing on the personal computer
Internet computing:-the computing of the mobile and computer system with the facility of the internet connectivity.

7 0

3 years ago

Translate the following MIPS code to C. Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3

Romashka [77]

Answer:

f = 2 * (&A[0])

See explaination for the details.

Explanation:

The registers $s0, $s1, $s2, $s3, and $s4 have values of the variables f, g, h, i, and j respectively. The register $s6 stores the base address of the array A and the register $s7 stores the base address of the array B. The given MIPS code can be converted into the C code as follows:

The first instruction addi $t0, $s6, 4 adding 4 to the base address of the array A and stores it into the register $t0.

Explanation:

If 4 is added to the base address of the array A, then it becomes the address of the second element of the array A i.e., &A[1] and address of A[1] is stored into the register $t0.

C statement:

$t0 = $s6 + 4

$t0 = &A[1]

The second instruction add $t1, $s6, $0 adding the value of the register $0 i.e., 32 0’s to the base address of the array A and stores the result into the register $t1.

Explanation:

Adding 32 0’s into the base address of the array A does not change the base address. The base address of the array i.e., &A[0] is stored into the register $t1.

C statement:

$t1 = $s6 + $0

$t1 = $s6

$t1 = &A[0]

The third instruction sw $t1, 0($t0) stores the value of the register $t1 into the memory address (0 + $t0).

Explanation:

The register $t0 has the address of the second element of the array A (A[1]) and adding 0 to this address will make it to point to the second element of the array i.e., A[1].

C statement:

($t0 + 0) = A[1]

A[1] = $t1

A[1] = &A[0]

The fourth instruction lw $t0, 0($t0) load the value at the address ($t0 + 0) into the register $t0.

Explanation:

The memory address ($t0 + 0) has the value stored at the address of the second element of the array i.e., A[1] and it is loaded into the register $t0.

C statement:

$t0 = ($t0 + 0)

$t0 = A[1]

$t0 = &A[0]

The fifth instruction add $s0, $t1, $t0 adds the value of the registers $t1 and $t0 and stores the result into the register $s0.

Explanation:

The register $s0 has the value of the variable f. The addition of the values stored in the regsters $t0 and $t1 will be assigned to the variable f.

C statement:

$s0 = $t1 + $t0

$s0 = &A[0] + &A[0]

f = 2 * (&A[0])

The final C code corresponding to the MIPS code will be f = 2 * (&A[0]) or f = 2 * A where A is the base address of the array.

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2 years ago

Magbigay ng tatlong hanap buhay na makukuha sa mineral produktong galing sa dagat at produktong agrikultural

OLEGan [10]

..............................................nice !

7 0

2 years ago

Read 2 more answers

Develop a CPP program to test is an array conforms heap ordered binary tree. This program read data from cin (console) and gives

mafiozo [28]

Answer:

Following are the code to this question:

#include<iostream>//import header file

using namespace std;

int main()//defining main method

{

int ar[7];//defining 1_D array that stores value

int i,x=0,l1=1,j; //defining integer variable

for(i=0;i<7;i++)//defining for loop for input value from user ends

{

cout<<"Enter a Number: ";//print message

cin>>ar[i];//input value in array

if(l1<=2 && i>0)//using if block that checks the array values

{

x++;//increment the value of x by 1

}

if(l1>2 && i>0)//using if block that checks the array values

{

l1=l1-2;//using l1 variable that decrases the l1 value by 2

}

j=i-x;//using j variable that holds the index of the root of the subtree

if(i>0 && ar[j]>ar[i])// use if block that checks heap condition

{

l1++; //increment the value of l1 variable

}

if(i>0 && ar[j]<ar[i])// using the if block that violate the heap rule

{

cout<<ar[i]<<" "<<"Violate the heap";//print message with value

break;//using break keyword

}

return 0;

}

Output:

Enter a Number: -15

Enter a Number: -5

-5 Violate the heap

Enter a Number: 45

Enter a Number: 0

Enter a Number: 55

55 Violate the heap

Explanation:

In the above-given C++ language code, an array "ar" and other integer variables " i,x,l1, j" is declared, in which "i" variable used in the loop for input values from the user end.
In this loop two, if block is defined, that checks the array values and in the first, if the block it will increment the value of x, and in the second if the block, it will decrease the l1 value by 2.
In the next step, j variable is used that is the index of the root of the subtree. In the next step, another if block is used, that checks heap condition, that increment the value of l1 variable. In the, if block it violate the heap rule and print its values.

5 0

2 years ago