In object-oriented programming, the object consist of both the data and the functions that operate on the data. In object-oriented programming, a data member plays the role of neither A nor B.
- Objects are the product of abstract data types that has inside of it both data and function altogether.
A data member often has different classes. They can be private or public.
Object-oriented programming (OOP) is simply known as a computer programming model that put together software design around data, or objects.
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Answer:
#Section 1
lst= []
lstNo=int(input("Enter even number of elements in List: "))
for i in range(0, lstNo):
pr=int(input(": " ))
lst.append(pr)
print(lst)
#Section 2
nlst=[]
dlst=[]
n =len(lst)
i=0
while (i < n/2):
nlst.append(lst[i])
i = i+1
j = n-1
while j >= n/2:
dlst.append(lst[j])
j = j-1
dlst.sort()
for a in range(len(dlst)):
nlst.append(dlst[a])
print(nlst)
Explanation:
#section 1:
An empty list is declared to hold the list inputs by the user <em>lst= []
</em>
The program then prompts the user to enter an even number of elements that will be contained in the list.
The for loop is used to iterate from zero(0) to the number of elements stated, in order to get the input that is appended to the list.
lastly, the list is printed out.
#Section 2:
In this section two new lists are created to hold one half of the value respectively.
The first list collects the first half of the list using a while loop and stores it, it does not perform any sorting on it.
The second list collects the second half using a while loop and sorts the list using the<em> .sort() </em>method which arranges elements in an ascending order.
Finally, the second list that has been sorted is added to the first and the result is printed to the screen.
A picture of how the code will run has been attached.
To transfer binary communication signals from source to destination
Answer:
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Explanation:
give me brainliest please
Answer:
The Anonymous Block will be:
1 DECLARE
2
3 lv_grade_row grade%rowtype;
4 lv_gr_avg grade.gr_t1%type;
5 lv_std_name varchar2(80);
6 cursor cursor_grade is select * from grade;
7
8 BEGIN
9
10 select avg(gr_t1) INTO lv_gr_avg from grade;
11 open cursor_grade;
12 fetch cursor_grade INTO lv_grade_row;
13
14 while cursor_grade%found loop
15
16 if
17 lv_grade_row.gr_t1>lv_gr_avg then lv_std_name :=
lv_grade_row.std_fname|| ' ' ||lv_grade_row.std_lname|| ' ' ||
lv_grade_row.gr_t1;
18 dbms_output.put_line(lv_std_name);
19
20 end if;
21 fetch cursor_grade into lv_grade_row;
22
Explanation: