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Temperature Conversion

 
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Temperature Conversion, Conversion chart
Tara200
13 Oct, 2007 - 11:46 AM
Post #1

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CODE
#include <iostream>

  using namespace std;


  int Cels = 0;
  float Faren = Cels * 9/5 + 32;
  char celsius = celsius;
  char farenheit = farenheit;
  
  int main ()
{
  
   float  Faren = Cels * 9/5 + 32;
   cout << " celsius   farenheit" << endl;
   cout << "--------------------" << endl;
   cout << "  0                 " << endl;
   cout << "  5                 " << endl;
   cout << " 10                 " << endl;
   cout << " 15                 " << endl;
   cout << " 20                 " << endl;
   cout << " 25                 " << endl;
   cout << " 30                 " << endl;

}


Hi, i am new to C++ and have to print out a chart that converts the celsius values above to their equivalent farenheit numbers. The print out has to be in the form of a chart with celsius in one column and farenheit in the other. My question is i am not sure how to convert the celsius numbers above in code as the code i have written here prints out the celsius numbers (given numbers) in a list with celsius at the head of it and farenheit prints out as a title in the next column. but i don't know how to get the celsius numbers to convert to farenheit and print out
in the rest of the chart. Any help would be greatly received as this is an assignment that has to be in this Thursday blink.gif

thankyou!
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Amadeus
RE: Temperature Conversion
13 Oct, 2007 - 12:38 PM
Post #2

g++ -o drink whiskey.cpp
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tempinF = (9/5)*tempinC + 32

Apply this formula for each value.

You need to input the Celsius temperatures to that formula, like so:

CODE

cout<<"  0                 "<<(9/5)*0 + 32<<endl;

and so on...

As an FYI, your program will not compile with your current variable declarations.

A more efficient way to complete the program would be to either place the Celsius temperatures in an array and loop through it, doing the conversions, or step through a loop with an incrementer of 5, as you seem to have a constant increment, and convert that.
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1lacca
RE: Temperature Conversion
13 Oct, 2007 - 12:45 PM
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code.rascal
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The "\t" (tabulator) escape character might help format the columns.
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gsewell
RE: Temperature Conversion
13 Oct, 2007 - 01:03 PM
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Dennis Ritchie, one of C's authors, addresses this subject in his exposition "The C Programming Language. Here is an example from section 1.2 which solves your mathematics as well as formatting issues:
CODE


#include <stdio.h>

main()
{
    float fahr, celsius;
    int lower, upper, step;

    lower = 0;
    upper = 300;
    step = 20;

    fahr = lower;
    while (fahr <= upper)
    {
        celsius = (5.0/9.0) * (fahr-32.0);
        printf("%3.0f %6.1f\n", fahr, celsius);
        fahr = fahr + step;
    }
}

Of course you can change the step and upper variables to suit your application.
-Gary
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Amadeus
RE: Temperature Conversion
13 Oct, 2007 - 01:08 PM
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g++ -o drink whiskey.cpp
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The code would also need to be translated to C++, which is of course easily done.

smile.gif
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Tara200
RE: Temperature Conversion
13 Oct, 2007 - 01:54 PM
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#include <iostream>

  using namespace std;


  int tempinC = 0;
  int tempinF = tempinC * (9/5) + 32;
  char celsius = celsius;
  char farenheit = farenheit;
  
  int main ()
{
  
  
   cout << " celsius   farenheit " << endl;
   cout << "-------------------- " << endl;
   cout << "  0         " <<  0 *  9/5 + 32 << endl;
   cout << "  5         " <<  5 *  9/5 + 32 << endl;
   cout << " 10         " << 10 *  9/5 + 32 << endl;
   cout << " 15         " << 15 *  9/5 + 32 << endl;
   cout << " 20         " << 20 *  9/5 + 32 << endl;
   cout << " 25         " << 25 *  9/5 + 32 << endl;
   cout << " 30         " << 30 *  9/5 + 32 << endl;
  
                

}

Thanks Very , Very much !! i am much obliged , i have adjusted it to the above code and it compiles and runs fine! smile.gif
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ahsesino
RE: Temperature Conversion
16 Oct, 2007 - 12:17 PM
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QUOTE(Tara200 @ 13 Oct, 2007 - 02:54 PM) *

CODE
#include <iostream>

  using namespace std;


  int tempinC = 0;
  int tempinF = tempinC * (9/5) + 32;
  char celsius = celsius;
  char farenheit = farenheit;
  
  int main ()
{
  
  
   cout << " celsius   farenheit " << endl;
   cout << "-------------------- " << endl;
   cout << "  0         " <<  0 *  9/5 + 32 << endl;
   cout << "  5         " <<  5 *  9/5 + 32 << endl;
   cout << " 10         " << 10 *  9/5 + 32 << endl;
   cout << " 15         " << 15 *  9/5 + 32 << endl;
   cout << " 20         " << 20 *  9/5 + 32 << endl;
   cout << " 25         " << 25 *  9/5 + 32 << endl;
   cout << " 30         " << 30 *  9/5 + 32 << endl;
  
                

}

Thanks Very , Very much !! i am much obliged , i have adjusted it to the above code and it compiles and runs fine! smile.gif



Well here is another way to doi with a loop hope it helps you.


#include <iostream>
using namespace std;

int main()
{
double celsius;
double fahrenheit;

cout << "Fahrenheit to Celsius conversion." << endl;
cout << "\nCelsius: Fahrenheit:\n";
cout << "--------------------------\n";


for (fahrenheit = 0; fahrenheit <= 30; fahrenheit++)
{




cout << fahrenheit << "\t\t " << fahrenheit * 9/5 + 32 << endl;
}

return 0;

}
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