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frozen [14]
3 years ago
12

Your task is to build a palindrome from an input string.A palindrome is a word that readsthe same backward or forward. Your code

will take the first 5 characters of the user input, and create a 9-character palindrome from it.Words shorter than 5 characters will result in a runtime error when you run your code. This is acceptablefor this exercise, however you already know how to validate user input with an IF statement.Some examplesof input words and the resulting palindromes:
Computers and Technology
1 answer:
White raven [17]3 years ago
7 0

Answer:

The program in Python is as follows:

word = input("Word: ")

if len(word) < 5:

   print("At least 5 characters")

else:

   pal = word[0:5]

   word = word[0:4]

   word = word[::-1]

   pal+=word

   print(pal)

Explanation:

This gets the word from the user

word = input("Word: ")

This checks if the length of the word is less than 5.

if len(word) < 5:

If yes, this tells the user that at least 5 characters is needed

   print("At least 5 characters")

If otherwise

else:

This extracts the first 5 characters of the word into variable named pal

   pal = word[0:5]

This extracts the first 5 characters of the word into variable named word

   word = word[0:4]

This reverses variable word

   word = word[::-1]

This concatenates pal and word

   pal+=word

This prints the generated palindrome

   print(pal)

You might be interested in
2- (8 point) Write a program using the instructions below. Assume that integers are stored in 4 bytes. a) Define an array of typ
lakkis [162]

Answer:

a)  

int apples [5] = {2, 4, 6, 8, 10};

b)

int *aPtr   //this is the pointer to int

Another way to attach a pointer to a an int variable that already exists:

int * aPtr;

int var;

aPtr = &var;

c)

for (int i = 0; i < size; i++){

       cout << values[i] << endl;    }

d)  

   aPtr = values;

   aPtr = &values[0];    

both  the statements are equivalent

e)

If its referring to the part d) then the address is:

cout<<aPtr;

f)

     for (int i = 0; i < size; ++i) {

            cout<<*(vPtr + i)<<endl;    }

g)

   cout << (aPtr + 3) << endl;  // address referenced by aPtr + 3

   cout << *(aPtr + 3) << endl; // value stored at that location

This value stored at location is 8

h)

    aPtr = &apples[4];

    aPtr -= 4;

    cout<<aPtr<<endl;

    cout<<*aPtr<<endl;  

Explanation:        

a)

int apples [5] = {2, 4, 6, 8, 10};

In this statement the array names is apples, the size of the array is specified in square brackets. so the size is 5. The type of array apples is int this means it can store integer elements. The values or elements of the array apples are even integers from 2 to 10. So the elements of array are:

apples[0] = 2

apples[1] = 4

apples[2] = 6

apples[3] = 8

apples[4] = 10

b)

In this statement int *aPtr  

The int* here is used to make the pointer aPtr points to integer object. Data type the pointer is pointing to is int. The asterisk symbol used with in makes this variable aPtr a pointer.

If there already exists an int type variable i.e. var and we want the pointer to point to that variable then declare an int type pointer aPtr and aPtr = &var; assigns the address of variable var to aPtr.

int * aPtr;

int var;

aPtr = &var;

c)

The complete program is:

int size= 5;

int values[size] = {2,4,6,8,10};

for (int i = 0; i < size; i++){

       cout << values[i] << endl; }

The size of array is 5. The name of array is values. The elements of array are 2,4,6,8,10.

To print each element of the values array using array subscript notation, the variable i is initialized to 0, because array index starts at 0. The cout statement inside body of loop prints the element at 0-th index i.e. the first element of values array at first iteration. Then i is incremented by 1 each time the loop iterates, and this loop continues to execute until the value of i get greater of equal to the size i.e. 5 of values array.

The output is:

2

4

6

8

10

d)

aPtr = values;

This statement assigns the first element in values array to pointer aPtr. Here values is the address of the first element of the array.

aPtr = &values[0];    

In this statement &values[0] is the starting address of the array values to which is assigned to aPtr. Note that the values[0] is the first element of the array values.

e)

Since &values[0] is the starting address of the array values to which is assigned to aPtr. So this address is the physical address of the starting of the array. If referring to the part d) then use this statement to print physical address is aPtr pointing to

cout<<aPtr;

This is basically the starting address of the array values to which is assigned to aPtr.

The output:

0x7fff697e1810                

f)

i variable represents offset and corresponds directly to the array index.

name of the pointer i.e. vPtr references the array

So the statement (vPtr + i) means pointer vPtr that references to array values plus the offset i array index that is to be referenced. This statement gives the address of i-th element of values array. In order to get the value of the i-th element of values array, dereference operator * is used.  It returns an ith value equivalent to the address the vPtr + i is pointing to. So the output is:

2

3

6

8

10

g)

values[0] is stored at 1002500

aPtr + 3 refers to values[3],

An integer is 4 bytes long,

So the address that is referenced by aPtr + 3 is

1002500 + 3 * 4 = 1002512

values[3] is basically the element of values array at 3rd index which is the 4th element of the array so the value stored at that referred location  is 8.

h)

Given that aPtr points to apples[4], so the address stored in aPtr is

1002500 + 4 * 4 = 1002516

aPtr -= 4  is equivalent to aPtr = aPtr - 4

The above statement decrements aPtr by 4 elements of apples array, so the new value is:

1002516 - 4 * 4 = 1002500

This is the address of first element of apples array i.e 2.

Now

cout<<aPtr<<endl; statement prints the address  referenced by aPtr -= 4 which is 1002500  

cout<<*aPtr<<endl;  statement prints the value is stored at that location which is 2.

6 0
4 years ago
What is the definition of Simplex transmission and Duplex Transmission?
ira [324]

Answer:

Alternatively referred to as simplex communication or simplex transmission, simplex is a one-way only communication standard that broadcasted information may only travel in one direction. This method is different than duplex transmission, which allows for two-way broadcasting. Examples of simplex include radio broadcasting, television broadcasting, computer to printer communication, and keyboard to computer connections.

Explanation:

computers

6 0
4 years ago
What is the purpose of fcc?
melamori03 [73]
<span> The Bureaus and the Office of Engineering and Technology process applications for licenses and other filings, analyze complaints, conduct investigations, develop and implement regulatory programs and participate in hearings, among other things. </span>
3 0
3 years ago
Pam is using the Outline view in PowerPoint. She would like to reposition
erastova [34]

Answer:

Explanation:

u have just to drag it into slide 6. I think that it's like this that we do it

6 0
3 years ago
Peter has recently bought a media player and a digital camera. He wants to buy a memory card for these devices. Which memory dev
lianna [129]

The answer is a Flash drive or better, an SD card

An SD card is a non-volatile memory use mostly in portable devices such as cameras, phones and other mobile solutions like tablets. They are capable of being a life in certain situations. They store all your documents, pictures, and photos in rows of tiny memory chips. In general, most modern consumer cameras use either Compact Flash or SD cards.


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