# double to float

• 11-07-2005
jsbeckton
double to float
I wrote a program converting fahrenheit to celsius but the requirement calls for using int for fahrenheit and float for celsius. Obviously thats leading to possible loss of data warnings. What can I do? And does anyone know why the "0" isn't showng in the celsius column? I think it was there but I changes something and now its gone? :confused:

Code:

```#include <stdio.h> #include <stdlib.h> int main() { int fahrenheit; float celsius; int lower, upper, step; printf( "Fahrenheit    Celsius\n\n" ); lower = 0; upper = 212; step = 1; fahrenheit = lower; while (fahrenheit <= upper)         {                 celsius = ( 5.0/9.0 ) * ( fahrenheit - 32.0 );                 printf( "%10.0d  %+10.3f\n", fahrenheit, celsius );                 fahrenheit = fahrenheit + step;         }         return 0; }```
• 11-07-2005
sunnypalsingh
Do these changes
Code:

```#include <stdlib.h> int main() { int fahrenheit; float celsius; int lower, upper, step; printf( "Fahrenheit    Celsius\n\n" ); lower = 0; upper = 212; step = 1; fahrenheit = lower; while (fahrenheit <= upper)         {                 celsius = ( 5.0f/9.0f ) * ( fahrenheit - 32.0f );                 printf( "%10.0d  %+10.3f\n", fahrenheit, celsius );                 fahrenheit = fahrenheit + step;         }         return 0; }```
• 11-07-2005
Alexthunder
Code:

```#include <stdio.h> #include <stdlib.h> int main() {   int fahrenheit;   float celsius;   int lower, upper, step;   printf( "Fahrenheit    Celsius\n\n" );   lower = 0;   upper = 212;   step = 1;   fahrenheit = lower;   while (fahrenheit <= upper)       {         celsius = ( 5.0f/9.0f ) * ( fahrenheit - 32.0f );         printf( "%10.0d  %+10.3f\n", fahrenheit, celsius );         fahrenheit = fahrenheit + step;       }   return 0; }```
ahihih