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C program to convert Fahrenheit to Celsius always prints zero
8 answers
Every day, a weather station receives 5 temperatures expressed in degree Fahrenheit. Write a program to accept temperature in Fahrenheit, convert it to Celsius and display the converted temperature on the screen. After 5 temperatures, the message ‘All temperatures processed’ should be displayed on the screen
My answer is below
#include <stdio.h>
main()
{
int fah, cel;
printf("\nEnter 5 temperatures in Fahrenheit: ");
scanf("%d %d %d %d %d", &fah);
do
{
cel=(fah-32)*(5/9);
printf("\n These temperatures converted to Celsius are: %d \n", cel);
}
while(fah);
}
scanf("%d %d %d %d %d", &fah);
You used 5 %d in the conversion specifier, but only one variable.
cel=(fah-32)*(5/9);
Here, you used integer division, change the type of the variables to double and 5/9 to 5.0/9
Lets take this one by one -
scanf("%d %d %d %d %d", &fah); : You want to scan five values while you are using only one %d. What about the other four values ? You should use an array of 5 values.cel=(fah-32)*(5/9); : You are dividing 5 by 9, both of which are integers. The result of dividing an integer by another, a/b, where a < b is 0. Hence cel will always be 0. You have to cast one of the integers to a float/double.- If you do use an array, you need to loop over all elements of the array to process them.
I am purposely not providing a direct solution to all these. I suggest you look up on these things and understand them, since they are very basic for programming in C.
You're working in integers --> 5/9 is converted to integer 0 thus the result will also equal to 0. You need to work with float/double types or change your calculation to minimize such rounding errors, e.g. something like this: cel=(5*(fah-32))/9; - here the multiplication is done first and division after that.
