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Harman [31]
3 years ago
13

The assignment operator has left-to-right-to-left associativity, which means that the value of the expression to the left of the

assignment operator is evaluated first and that the result is assigned to the operator on the right. True or false?
Computers and Technology
1 answer:
postnew [5]3 years ago
4 0

Answer:

The given statement is false statement.

Explanation .

  • The assignment is assigning the value to the variable or we can say that it will also be used to initialize the variable The "=" is the symbol of the assignment operator.
  • For example

        int r=90, The value of variable 90 is assigned to 90.

  • The associativity, of assignment, is always right to left which means The right-hand side is firstly evaluated then it will be assigned in left-hand side variable or expression.  

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Template files can contain the following items: text, tables, images and lists.
TEA [102]

It is true that template files can contain items such as text, tables, images and lists

Explanation:

A template is a pre-designed document file that is formatted within a default layout and is used to help users create new documents such as letters, tables, resumes, presentations, and reports. A resume template, for instance, already contains placeholder text that users can use to replace text information of their own. It would most likely have a blank space for details like your name, address, and contact information. Some programs have templates that are pre-built while others can be created from scratch. Microsoft PowerPoint has design templates that contain visual representation of images as well. These images help achieve document cohesion.

Learn more about templates

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#LearnWithBrainly

4 0
3 years ago
When the tcp\ip translates a network layer address into a data link layer address (after not finding an entry for the network la
Pavel [41]
When the network stack translates an IP address (network address) to a MAC address (data link address) it sends a broadcast using ARP to all devices on the subnet asking who has the specified IP address.  The device with that IP address will then respond with it's MAC address to the ARP message's source.

With this in mind the answer to your question would either be broadcast or ARP request depending on what you have been learning in class.
6 0
2 years ago
Write a function called count_occurrences that takes two strings. The second string should only be one character long. The funct
iVinArrow [24]

Answer:

Here is the Python function:

def count_occurrences (string1 , string2):  # method that takes two strings as parameter and returns how many times second string occurs in first

   count = 0  #counts number of occurrence of string2 in string1

   for word in string1:  #iterates through each word of the string1

       for character in word:  #iterates through each character of each word in string1

           if character == string2:  # checks if the character in string1 is equal to the character of string2

               count = count + 1  #adds 1 to the count of string2 in string1

   return count  #returns number of times the string2 occurs in string1

#in order to check the working of the function add the following lines to the code:    

first_str = input("Enter the first string: ")  #prompts user to enter first string

second_str = input("Enter the second string (one character long): ") #prompts user to enter second string which should be one character long

occurrence = count_occurrences(first_str,second_str)  #calls count_occurrences method by passing both strings to it

print(second_str,"occurs",occurrence,"times in",first_str) #prints how many times the second_str occurs in the first_str on output screen

Explanation:

The program is well explained in the comments added to each statement of the above code. I will explain the program with an example:

Suppose

string1 = "hello world"

string2 = 'l'

So the method count_occurrences() should return how many times l occurs in hello world.

count is used to count the number of occurrences of l in hello world. It is initialized to 0.

The first loop iterates through each word of string1. The first word of string1 is "hello"

The second loop iterates through each character/letter of each word. So the first letter of "hello" word is 'h'.

The if condition if character == string2: checks if the character is equal to string2 i.e. l.

character = "h"

string2 = "l"

The are not equal so the program moves to the next character of word hello which is e. The if condition again evaluates to false because e is not equal to l. So the program moves to the next character of word hello which is l.

This time the if condition evaluates to true because

character = "l"

string2 = "l"

character == string2

l == l

So the if part is executed which has the following statement:

count = count + 1

So the count is added to 1. count was 0 previously. Now

count = 1

Next the program moves to the next character of word hello which is l. This time the if condition evaluates to true because the character l matches with string2. So count is again incremented to 1. count = 2.

Next the program moves to the next character of word hello which is o. The if condition evaluates to false as o does not match with string2 i.e. l. So the count remains 2.

Next the first loop moves to the next word of the string1 = "world". The inner (second) for loop iterates through each character of "world" from "w" to "d" and checks if any character is equal to string2. Only one character is equal to string2 in "world". So count is incremented to 1. Hence count = 3

After the loop ends the statement return count returns the number of times string2 occurs in the string1. So the output is:

l occurs 3 times in hello world  

8 0
3 years ago
Write a program that uses the function isPalindrome given in Example 6-6 (Palindrome). Test your program on the following string
Artist 52 [7]

Answer:

#include <bits/stdc++.h>

using namespace std;

bool isPalindrome(string str)

{

   char a,b;

int length = str.length();

for (int i = 0; i < length / 2; i++)

{

   a=tolower(str[i]);//Converting both first characters to lowercase..

   b=tolower(str[length-1-i]);

   if (b != a )

return false;

}

return true;

   

}

int main() {

   string t1;

   cin>>t1;

   if(isPalindrome(t1))

   cout<<"The string is Palindrome"<<endl;

   else

   cout<<"The string is not Palindrome"<<endl;

return 0;

}

Output:-

Enter the string

madam

The string is Palindrome

Enter the string

abba

The string is Palindrome

Enter the string

22

The string is Palindrome

Enter the string

67876

The string is Palindrome

Enter the string

444244

The string is not Palindrome

Explanation:

To ignore the cases of uppercase and lower case i have converted every character to lowercase then checking each character.You can convert to uppercase also that will also work.

6 0
2 years ago
Match the cell reference to its definition
Alisiya [41]

Answer:

which cell reference:-|

6 0
3 years ago
Read 2 more answers
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