Answer:
Let the customer vent and then restart the conversation
Explanation:
It is best for Derek in this situation allow the customer say everything on his or her mind about the problems at hand. This will help Derek identify and understand the problems and fix them accordingly. However, he'd need to start the world all over again to achieve this.
Cheers
Answer:
Boundary folding method is basically used in the java algorithm and in the hash table. In the hash function, the left and the right value are basically folded in the fixed boundary between the given center values by using the boundary folding methods.
There are basically two types of folding method in the hashing that are:
- Folding shift
- Folding boundary
In the folding boundary method the outside value are get reversed and the alternate values are get flipped at the boundary folding method.
They may invade privacy rights of air space and be used for other unintended acts such a bombings and espionage
Answer:
\n
Explanation:
readline() method is used to read one line from a file. It returns that line from the file.
This line from the file is returned as a string. This string contains a \n at the end which is called a new line character.
So the readline method reads text until an end of line symbol is encountered, and this end of line character is represented by \n.
For example if the file "abc.txt" contains the lines:
Welcome to abc file.
This file is for demonstrating how read line works.
Consider the following code:
f = open("abc.txt", "r") #opens the file in read mode
print(f.readline()) # read one line from file and displays it
The output is:
Welcome to abc file.
The readline() method reads one line and the print method displays that line.
Answer:
#include <iostream>
#include <string>
#include <stack>
#include <math.h>
using namespace std;
int main() {
string s;
double n=0;
int position=0;
stack<int> wholeNumbers;
cout<<"Enter a decimal number:";
cin>>s;
string::iterator counter = s.begin();
while(*counter!='.' && counter!=s.end()){
wholeNumbers.push(*counter-48);
counter++;
position=position+1;
}
for(int i=0;i<position;i++){
n=n+(wholeNumbers.top()*pow(10,i));
wholeNumbers.pop();
}
position=-1;
if(counter!=s.end()){
counter++;
}
while(counter!=s.end()){
n=n+((*counter-48)*pow(10,position));
position=position-1;
counter++;
}
cout<<n;
}
Explanation:
- Inside the while loop, push the push a number to the wholeNumbers stack by subtracting it with 48.
- Increment the counter and position variable by 1 inside the while loop.
- Count the number of digit, push each digit to top of stack and find the end of the number,
- Run a for loop up to the value of position variable and pop a value from the wholeNumbers stack.
