Celsius-Fahreneit program

[cogeek]

I got the following code i written for an exercise of an a book

#include <stdio.h>

main()
{
int ce;
int fa;
printf("Write fahreneit: ");
scanf("%i", &fa);
ce= (5\9)*(fa-32);
printf("The celsius are %i\n", ce);
}

i don't get any error when compiling vut when running i get
The celsius are 0
i get this by writing differnet numbers, only 0, why does this happens?

[Brain Cell]

- click here and here

- remember the right division operator


there you go , your program works now..


two more things :

- COMPILE the code before you put it here.

- READ THE GOD DAMN ERRORS YOUR COMPILER SHOW !!!!!!! its like you're saying "there , YOU compile my code , YOU read the errors and YOU fix it"

[cogeek]

i don't get any error when compiling

i don't get any erros

[Brain Cell]

How would it compile with an error like that? :

ce= (5\9)*(fa-32);

both VC++ and Dev C++ shows that error and wouldn't compile before you fix it

whats your compiler anyway?

[hk_mp5kpdw]

Both the formula in the OP for converting fahrenheit to celsius and the link provided by Brain Cell are correct. The OP uses C=(5/9)*(F-32). The link uses C=(F-32)/9*5. These are essentially the same thing (one form happens to be better than the other for purposes of programming however), so...

Your problem comes from trying to do the integer division of 5 by 9 (5/9). This will always be 0 and your equation therefore becomes:

ce= 0*(fa-32);

Which I hope you can see will also always be 0 no matter what fa happens to be. 0 times anything is still 0.

Use floating point variables (float or double) to store your values and do your computations with, i.e.:

float ce;
float fa;
printf("Write fahrenheit: ");
scanf("%f", &fa);
ce= (5.0f/9.0f)*(fa-32);
printf("The celsius is %f\n", ce);

[cogeek]

ok,
i get it:
whats your compiler anyway?
my compiler is gcc!

[sand_man]

turn the wanrings on gcc

[goosematt]

I got the following code i written for an exercise of an a book

#include <stdio.h>

main()
{
int ce;
int fa;
printf("Write fahreneit: ");
scanf("%i", &fa);
ce= (5\9)*(fa-32);
printf("The celsius are %i\n", ce);
}

i don't get any error when compiling vut when running i get
The celsius are 0
i get this by writing differnet numbers, only 0, why does this happens?

gcc -Wall -o file file.c (thats a good start (Wall means turn all warnings on) this is what i get:

guyattc2-practical_c3-> gcc -Wall -o file file.c
file.c:4: warning: return type defaults to `int'
file.c: In function `main':
file.c:9: error: stray '\' in program
file.c:9: error: syntax error before numeric constant

for a start helps to assign main() to an int as this is what this default function returns. To exit a program properly is exits by the statement:

return(0);

also errors already spotted: you need to use floats and also divide is '/' not '\'. Also, a 'nicer' way that ive been told to use with user input is to use fgets and sscanf as well. This is how my book explains it (im a newbie too ). scanf is described as 'dodgy'. So my friend, here is my code:

#include <stdio.h>
#include <string.h>

int main()
{

  float celsius;     /* celsius */
  float fahrenheit;  /* fahrenheit */
  char  line[50];    /* a line to take user input */
  
  printf("Write fahreneit: ");
  
  fgets(line, sizeof(line), stdin); /* take user input and store in 'line' */
  
  sscanf(line, "%f", &fahrenheit); /* take the input and find the float
                                       ("%f") and store this value in
				       variable fahrenheit
				    */
  
  celsius = (5.0f/9.0f) * (fahrenheit - 32);
  
  printf("Fahrenheit %f == Celsius %f\n", fahrenheit, celsius);
  
  return(0); /* return '0' to exit program normally */
}

With this code, i get this on the command:

guyattc2-practical_c3-> gcc -Wall -o file file.c
guyattc2-practical_c3-> ./file
Write fahreneit: 32
Fahrenheit 32.000000 == Celsius 0.000000

hope this solves lots of problems!

--CHriS

[cogeek]

thanks,
nice example