Determining if two numbers are "almost equal" and outputting the result - c++

Question

Please note that I am just starting C++.
I am working from Bjarne Stroustrup's book "Programming Principles and Practice Using C++ 2nd Edition" and am on a Drill at the end of chapter 4.
These are the instructions so far:
Write a program that consists of a while-loop that (each time around the loop) reads in two ints and then prints them. Exit the program when a terminating '|' is entered.
Change the program to write out "the smaller value is:" followed by the smaller of the numbers and the "larger value is:" followed by the larger value.
Augment the program so that it writes the line "the numbers are equal" (only) if they are equal.
Change the program so that it uses doubles instead of ints.
5. Change the program so that it writes out "the numbers are almost equal" after writing out which is the larger and the smaller if the two numbers differ by less than 1.0/100.
This is where I am stuck.
First, I do not understand what "if the two numbers differ by less than 1.0/100." means. Does that mean if the two numbers are within 100 numbers of each other?
Second, how do I determine "if the two numbers differ by less than 1.0/100"? (The more simple the solution the better.)
Here is my code so far:
#include "std_lib_facilities.h"
int main()
{
double v1 = 0, v2 = 0;
cout << "Enter two numbers: ";
while(cin >> v1 >> v2)
{
if (v1 > v2)
cout << "The smaller of the two numbers is: " << v2 << "\n";
else if (v1 == v2)
cout << "The numbers are equal. \n";
else
cout << "The smaller of the two numbers is: " << v1 << "\n";
cout << "Enter two numbers: ";
}
}
Thank you for taking the time to read this.

Answer 1

This may be what it means
if(std::abs(v1-v2)<0.01)
#Walter 's answer may also be correct. It depends on the intention of the original question, which is not quite clear. For example, if #Walter 's version is used, then for v1 = v2 = 0.0, the program would say they are not close.

Answer 2

I think it means that the relative difference between the two floating point numbers is less than 1 part in 100, i.e.
if(std::abs(x-y)<0.01*std::max(std::abs(x),std::abs(y)))
Here I use the maximum of the absolute values as reference. You may also use the mean, but taking (the absolute of) either value is less advisable because it's not symmetric.

Answer 3

I have understood the assignment the following way:
#include "std_lib_facilities.h"
int main()
{
double v1 = 0, v2 = 0;
cout << "Enter two numbers: ";
while(cin >> v1 >> v2)
{
if ( v1 < v2 )
{
cout << "The smaller of the two numbers is the first number: " << v1 << "\n";
cout << "The largest of the two numbers is the second number: " << v2 << "\n";
}
else if ( v2 < v1 )
{
cout << "The smaller of the two numbers is the second number: " << v2 << "\n";
cout << "The largest of the two numbers is the first number: " << v1 << "\n";
}
if ( abs( v1 - v2) < 1.0 / 100 )
{
cout << "The numbers are almost equal. \n";
}
cout << "Enter two numbers: ";
}
}

Answer 4

This is a dead topic but I thought I should clarify this. The difference between 2 numbers is the result of the subtraction between the largest number and the smallest number. Therefore you need to declare 2 variables to keep track of which is the smallest and which is the largest in each of the if-else statements. Therefore the code should look like this.
#include "std_lib_facilities.h"
int main()
{
double v1 = 0, v2 = 0, larger, smaller;
cout << "Enter two numbers: ";
while(cin >> v1 >> v2)
{
if (v1 > v2)
{
cout << "The smaller of the two numbers is: " << v2 << "\n";
smaller = v2;
larger = v1;
}
else if (v1 == v2)
{
cout << "The numbers are equal. \n";
}
else if(v1 < v2)
{
cout << "The smaller of the two numbers is: " << v1 << "\n";
smaller = v1;
larger = v2;
}
if((larger-smaller)<1.0/100)
{
cout << "the numbers are almost equal" << endl;
}
cout << "Enter two numbers: ";
}
}
return 0;

Answer 5

This is my version:
#include "std_lib_facilities.h" // Custom header file for this book
int main() {
double a, b;
while (cin >> a >> b) {
if (a < b) {
cout << "The smaller value is: " << a << '\n'
<< "The larger value is: " << b << '\n';
if ((b - a) < (1.0 / 100))
cout << "The numbers are almost equal\n";
}
else if (a > b) {
cout << "The smaller value is: " << b << '\n'
<< "The larger value is: " << a << '\n';
if ((a - b) < (1.0 / 100))
cout << "The numbers are almost equal\n";
}
else {
cout << "The numbers are equal\n";
}
}
return 0;
}

Answer 6

#include <iostream>
#include <string>
using namespace std;
int main(){
double n1, n2;
double smaller, larger;
double differ=1.0/100;
string small ="The smallest number is: ";
string large ="The largest number is: ";
string equal ="\nBoth numbers are almost equal\n";
while(cin>>n1>>n2){
if (n1==n2){
cout<<"Both number are abslute equal!!!";
continue;
}
else if(n1>n2){
cout<<small<<n2<<" ";smaller=n2;larger=n1;}
else if (n1<n2) {
cout<<large<<n2<<" ";larger=n2;smaller=n1;}
if(larger-smaller<=differ)
cout<<equal;
}
return 0;
}

Answer 7

Even though it's an old thread I've come across it whilst working through the the same book. In case any more students do the same, here's what I've come up with using 1.0 / 100 as the literal 0.01:
#include "../../std_lib_facilities.h"
double lower(double int1, double int2) //function to calculate the smallest of two doubles
{
if (int1 > int2)
return int2;
else
return int1;
}
double upper(double int1, double int2) //function to calculate the largest of two doubles
{
if (int1 < int2)
return int2;
else
return int1;
}
int main()
{
double val1, val2;
while (cin >> val1 >> val2)
{
double smallest = lower(val1, val2);
double largest = upper(val1, val2);
if (val1 == val2)
{
cout << "The values are equal.\n";
}
else
{
cout << "The smaller value is: " << smallest << "\n";
cout << "The larger value is: " << largest << "\n";
if (abs(largest - smallest) < 1.0 / 100)
{
cout << "The numbers are almost equal.\n";
}
}
}
}
By this point in the book you too should have already covered functions and can utilise them here to keep your main function tidy.

Answer 8

"5. Change the program so that it writes out "the numbers are almost equal" after writing out which is the larger and the smaller if the two numbers differ by less than 1.0/100."
The reason why you do not understand what the author is asking is because the question is based on SPECIFICALLY pure basic mathematical speech (Was hard for me to understand "math" word questions too). It's one of those questions you can only understand if you understand the grammar of math, complete the book: "Practical Algebra: A Self-Teaching Guide" finish that book and then read c++ from Bjarne, it's going to be extremely difficult if you can't understand basic mathematical grammar specially for hardcore programming language c++.
"the number are almost equal"
It is a figure of speech, doesn't mean that there's a special C++ function or library for it.
"if the two numbers differ by less than 1.0/100."
Literally means the difference of two numbers are equal to 1.0/100 which is 0.01, which means for example if (9.99 - 10) = 0.01. But there's more to the question which makes it tricky, not only the Author is asking for the difference of 0.01 but the difference has to be less than 0.01. so basically you can write the formula such as:
double a; // use the built-in type keyword "double" since the values are in decimal.
double b;
if ((a - b) < 0.01)
{
//statement code goes here.
}
Also you deserve a cookie since you're reading a C++ book from it's creator.

Answer 9

This has been working for me.
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
double x=0;
double y=0;
while(cin >> x ,cin >> y)
if (x > y){
cout << "The larger value is " << x << " the smaller value is " << y << "\n";
if (x - y < (1.0/100)) {
cout << "the numbers are almost equal" << endl;
}
}
else if (x < y){
cout << "The larger value is " << y << " the smaller value is " << x << "\n";
if (y - x < (1.0/100)) {
cout << "the numbers are almost equal" << endl;
}
}
else if (x == y)
cout << "the numbers must be equal\n";
}