When using the strcat function, you must be careful not to overwrite the bounds of an array?

Suppose we have two String objects and treat the characters in each string from beginning to end in the following way: With one

Simora [160]

Answer:

If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.

Explanation:

Lets take an example of two strings abc and cba which are reverse of each other.

string1 = abc

string2 = cba

Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.

Push abc each character on a stack in the following order.

c

b

a

Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.

Insert cba each character on a stack in the following order.

a b c

First c is added to queue then b and then a.

Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.

First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match

c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.

6 0

3 years ago

Complete the function to return the result of the conversiondef convert_distance(miles):km = miles * 1.6 # approximately 1.6 km

AnnZ [28]

Answer:

The complete program is as follows:

def convert_distance(miles):

km = miles * 1.6 # approximately 1.6 km in 1 mile

return km

my_trip_miles = 55

# 2) Convert my_trip_miles to kilometers by calling the function above

my_trip_km =convert_distance(my_trip_miles) #3) Fill in the blank to print the result of the conversion

# 4) Calculate the round-trip in kilometers by doubling the result,

print("The distance in kilometers is " +str(my_trip_km))

# and fill in the blank to print the result

print("The round-trip in kilometers is " + str(my_trip_km * 2))

Explanation:

The program is self-explanatory because I used the same comments in the original question.

5 0

2 years ago

List the six external parts or "peripherals" of a computer system and identify which are output and which are input devices/ Lis

vitfil [10]

Keyboard-Input
Mouse-Input
Monitor-Output
Speakers-Output
Printer-Output
Hard Drive-Output

3 0

3 years ago

Which of the following statements is the least abstraction of the world wide web

Sliva [168]

Documents, images and other data you can access by providing a uniform Resource Locator. URL - the Web Address

I hope this helps. You didn't give me the choices to from.

8 0

3 years ago

Read 2 more answers

⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val

timurjin [86]

Answer:

a) There are approximately 2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b) ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 = 4.824 × 10^9 bytes. Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a) There are approximately 2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b) ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 = 4.824 × 10^9 bytes. Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

8 0

3 years ago