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disa [49]
3 years ago
13

Given variables first and last, each of which is associated with a str, representing a first and a last name, respectively. Writ

e an expression whose value is a str that is a full name of the form "Last, First". So, if first were associated with "alan" and last with "turing", then your expression would be "Turing,Alan". (Note the capitalization! Note: no spaces!) And if first and last were "Florean" and "fortescue" respectively, then your expression's value would be "Fortescue,Florean".

Computers and Technology
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

The python code is attached

Explanation:

  1. I defined a function called Fullname
  2. The function accepts 2 parameters first and last
  3. last[0] gives the first letter of the last variable
  4. last[0].upper() modifies the first letter as upper letter
  5. same applied to variable first
  6. + sign concatenates strings and variables to the variable name
  7. the function returns the variable name

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Can somebody explain me what this code does in a few or one sentence?#include #include using namespace std;int main () { const i
goblinko [34]

Answer:

The program declares and array of 8 elements and swap the position of the 1st element with the 8th; the 2nd with the 7th, the 3rd with the 6th and the 4th with the 5th element

Explanation:

First, I'll arrange the code line by line, then I'll start my explanation from the variable declaration at line 4

#include

using namespace std;

int main () {

const int NUM_ELEMENTS = 8;

vector numbers(NUM_ELEMENTS);

int i = 0;

int tmpValue = 0;

cout << "Enter " << NUM_ELEMENTS << " integer values..." << endl;

for (i = 0; i < NUM_ELEMENTS; ++i)

{

cout << "Enter Value#" << i+1 << ": "; cin >> numbers.at(i);

}

for (i = 0; i < (NUM_ELEMENTS /2); ++i)

{

tmpValue = numbers.at(i); numbers.at(i) = numbers.at(NUM_ELEMENTS - 1 - i);

numbers.at(NUM_ELEMENTS - 1 - i) = tmpValue;

}

system ("pause");

return 0;

}

Line 4: This line declares an instant constant variable NUM_ELEMENTS with a constant value of 8. Meaning that the value cannot be changed during program execution.

Line 5: This line declares a vector array that can change in size. Here, it was declared with size 8.

Line 6 & 7: These lines declares integers variables i and tmpValue with an initialised value of 0, each.

Line 8: This line prints the following string; Enter 8 integer values...

Line 9 to 12: These lines represent an iteration which starts from 0 to 7.

Side Note: When an array is being declared, the index starts from 0 and ends at 1 less that the array size.

So, during this iteration, it accepts inputs into the array from index 0 to 7 i.e. from the first element till the last

Line 13 to 17: This is also an iterative statement. But what this iteration does is that, it swaps elements of the array (as stated in the answer section)

The iteration starts from 0 and ends at a value less than NUM_ELEMENTS/2

Note that NUM_ELEMENTS = 2

So,we can conclude that the iteration starts from 0 till a value less that 8/2

Iteration: 0 till a value less that 4

So, that's 0 to 3 (the iteration is done on array element at index 0,1,2 and 3).

When iteration is at 0, the following is done

tmpValue = number at index 0 i.e. a temporary value is used to store the number at index 0 of the array

Number at 0 = number at (8-1-0)

i.e. number at 0 = number at (7)

Number at 7 is then equal to tmpValue

Swap Completed.

The same is done for index 1,2 and 3.

4 0
3 years ago
Please explain this code line by line and how the values of each variable changes as you go down the code.
Scilla [17]

Answer:

hope this helps. I am also a learner like you. Please cross check my explanation.

Explanation:

#include

#include

using namespace std;

int main()

{

int a[ ] = {0, 0, 0};  //array declared initializing a0=0, a1=0, a3=0

int* p = &a[1]; //pointer p is initialized it will be holding the address of a1 which means when p will be called it will point to whatever is present at the address a1, right now it hold 0.

int* q = &a[0];  //pointer q is initialized it will be holding the address of a0 which means when q will be called it will point to whatever is present at the address a0, right now it hold 0.

q=p; // now q is also pointing towards what p is pointing both holds the same address that is &a[1]

*q=1 ; //&a[0] gets overwritten and now pointer q has integer 1......i am not sure abut this one

p = a; //p is now holding address of complete array a

*p=1; // a gets overwritten and now pointer q has integer 1......i am not sure abut this one  

int*& r = p; //not sure

int** s = &q; s is a double pointer means it has more capacity of storage than single pointer and is now holding address of q

r = *s + 1; //not sure

s= &r; //explained above

**s = 1; //explained above

return 0;

}

6 0
3 years ago
Discuss, in detail, the usefulness of graphics and images in program development. Explain the importance of the interface of a c
vesna_86 [32]

Answer:

Explanation:

The main importance of graphics and images in program development is user experience. These graphics, color schemes, clarity, images etc. all come together to create a user interface that individuals will be able to easily navigate and use with extreme ease in order to benefit from what the program is intended to do. An interface of a class of objects allows every button and object in the program to grab and share functionality in order for the entire program to run smoothly.

4 0
3 years ago
Jeremy is working with a team that is creating an application using attributes and associated methods. What type of programming
Ugo [173]
<span>object-oriented programming languages</span>
8 0
3 years ago
Justine was interested in learning how to play the piano. Before she was allowed to even play the piano, she has had to learn
Setler [38]

Answer:

A

Explanation:

The answer is A because, to be novice at something that means one is new to and is inexperienced at said activity.

3 0
3 years ago
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