Celsius to Fahrenheit with while loop

[thaooherb]

I write the code for this problem but it doesn't work. Can anyone provide me some idea?

Write a program that finds the temperature, as an integer, that is the same in both Celsius and Fahrenheit. The formula to convert from Celsius to Fahrenheir is:

Fahrenheit = 9/5 Celsius + 32

Your program should create two integers variables for the temperature in Celsius and Fahrenheit. Initialize the temperature to 100 degrees Celsius. In a loop, decrement the Celsius value and compute the corresponding temperature in Fahrenheit until the two values are the same.

  #include <iostream>

using namespace std;

int main()
{
    double C=100, F;
    while (C--<101)
    {
        F=(9/5)*C + 32;
        if (F==C)
            break;

    }
    cout<<"The same temperature is "<<C;
    return 0;
}

[Chervil]

I'm not sure this loop condition is very useful:

    while (C--<101)

It should loop forever, as C starts less than 101 and never increases. You may as well put while (true) which explicitly says you want to loop forever.

This line always gives problems. I suspect it is one of the main points of this exercise.

F=(9/5)*C + 32;

9 is an integer. 5 is an integer. 9/5 does integer division resulting in 1. The effect is F = C + 32;. And that means the following condition, if (F==C) will never be true. Well, comparing floating-point types is a separate topic. I won't go too deep into that here.

This line should be

F = (9.0/5.0)*C + 32;

which will cause the numbers to be type double, giving 9.0/5.0 == 1.8.


edit: original instructions state:

Your program should create two integers variables for the temperature in Celsius and Fahrenheit.

Hence F and C should be int, not double.
But you still need 9.0/5.0 as type double.

[gunnerfunner]

# include <iostream>

auto CelToFah(const int tempCel)
{
    return (1.8 * tempCel) + 32;//see Chervil's comments above 
}

int main()
{
    int tempCel = 100;

    while (static_cast<int>(CelToFah(tempCel)) != tempCel)
    {
          --tempCel;
    }
    std::cout << "C == F @ " << tempCel << "\n";
}

edit: the problem pre-supposes that there is indeed a temp at which C == F but a robust program should also cover the possibility that there might not be one and stop checking at a predifined temp (in C), say -10'000 etc and report that no matches were found over given range etc

[JLBorges]

> stop checking at a predifined temp (in C), say -10'000

See: https://www.britannica.com/science/absolute-zero

[gunnerfunner]

yeah, that was a clanger on my part. thanks,

[lastchance]

It's a poor question.

There is no a priori reason for a solution in integers.

F = 1.8 C + 32 necessitates truncation 80% of the time and I'm not sure whether cast'ing (of any form) will necessarily produce the nearest integer for both positive and negative values of the RHS.

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
   int C = 100, F;
   do
   {
      C--;
      F = round( 1.8 * C + 32 );
      cout << C << '\t' << F << '\n';
   } while ( F > C );       // initial values and gradient ensure this must eventually end
   cout << "The same temperature is " << C << '\n';
}

[JLBorges]

> It's a poor question.
> There is no a priori reason for a solution in integers.
> F = 1.8 C + 32 necessitates truncation 80% of the time ...

There is nothing wrong with the original question which stated (using integer arithmetic):
"The formula to convert from Celsius to Fahrenheit is: Fahrenheit = 9/5 Celsius + 32"

This equation ( x*5 = x*9 + 160 ) is guaranteed to have an integer solution (less than zero).

What are poor are the programs which attempted to compare the result of floating point calculations.

#include <iostream>

int main()
{
    // const int SMALLEST_INT_CELSIUS = -273 ;

    int celsius = 100 ; // Initialize the temperature to 100 degrees Celsius

    do // in a loop
    {
        --celsius ; // decrement the Celsius value

        /*
        if( celsius < SMALLEST_INT_CELSIUS )
        {
            // report error: solution to the equation is below absolute zero
        }
        */

        // and compute the corresponding temperature in Fahrenheit
        const double fahrenheit = celsius * 9.0 / 5 + 32 ;

        // presumably print out the values
        std::cout << "Celsius " << celsius << " == " << fahrenheit << " Fahrenheit\n" ;

    } while( celsius != ( celsius * 9 / 5 + 32 ) ) ; // until the two values are the same
}

[lastchance]

The only reason that we know it has an integer solution is because we have already solved it by simple algebra and found it thus.

The original question states "Your program should create two integers variables for the temperatures in Celsius and Fahrenheit ... "

[JLBorges]

> The only reason that we know it has an integer solution
> is because we have already solved it by simple algebra and found it thus.

Repeat:
For any equation a.x = b.x + c, where x is the variable and a, b, and c are (edit: finite) non-zero integers,
we do not have to solve the equation to know that it has an integer solution.


> The original question states "Your program should create two integers variables for the
> temperatures in Celsius and Fahrenheit ... "

To determine the solution to such an equation in the program, it is required that we use some kind of integer variables for x and y. To ask the programmer to use non-integer variables for x and y would be downright silly.

[lastchance]

The equation a.x = b.x + c
has solution x = c/(a-b)

This is an integer only if a-b is a factor of c. This is the case here as -4 divides a whole number of times into 160. It is not true for arbitrary a, b, c.

If the equation had been F = (9/5) C + 31
then there would have been no integer solution. Try it.

[JLBorges]

> The equation a.x = b.x + c has solution x = c/(a-b)
> This is an integer only if a-b is a factor of c. It is not true for arbitrary a, b, c.

Right; I stand corrected.
Thank you!

[thaooherb]

Thank you guys. I figured out the problem. This is my code.

#include <iostream>

using namespace std;

int main()
{
    int C=100, F;
    F=(9*C)/5 + 32;
    while (F!=C)
    {
        C--;
        F=(9*C)/5 + 32;
    }
    cout<<"The same temperature is "<<F;
    return 0;
}