Issue with fahrenheit conversion formula in C [duplicate]

Question

When writing a program in C to convert celsius to fahrenheit, the following formula gives the incorrect output:

    int fahr = 9 / 5 * celsius + 32;

Now, I understand that this is probably an issue with 9/5 being interpreted as an integer, but what I don't understand is that using double or floatit still gives the same incorrect output.

Oddly enough the following formula gives the correct output despite also setting the type to int:

int fahr = celsius / 5 * 9 + 32;

Furthermore, i've noticed even something as simple as the below, when the type is set to double, still gives the output as 1.0 instead of 1.8:

 double x = 9 / 5;
 printf("%lf\n", x);

I've read this thread:

C program to convert Fahrenheit to Celsius

but I still don't understand why int fahr = celsius / 5 * 9 + 32; works but not int fahr = 9/5 * celsius+32; ?

Answer 1

You're doing math with integers. This expression:

9 / 5

Yields 1 in C. When you say you used double or float, you probably just changed the type of fahr, which doesn't do anything to the operations taking place on the right side of the assignemtn operator. To get the right behaviour, you need to make at least one of those constants a double, too:

9.0 / 5

Likewise, in this statement:

double x = 9 / 5;

You're still doing integer math, and then assigning the result to a double variable. There isn't anything else going on. You'll get the right answer by doing one of these:

double x = 9.0 / 5;
double x = 9 / 5.0;
double x = 9.0 / 5.0;

The reason this expression:

int fahr = celsius / 5 * 9 + 32;

appears to work is just an order of operations thing - here you divide the input by 5 and then multiply by nine, rather than doing the constant operation first. You'd still get more accurate answers by doing:

int fahr = celsius * 9 / 5 + 32;

Besides that, you could also do floating point math in this expression:

int fahr = celsius * 9.0 / 5 + 32;

If you want to do the original calculation using integers, you certainly can - you just need to multiply before dividing:

int fahr = 9 * celsius / 5 + 32;

This expression is equivalent to one of the ones used above.

Answer 2

I still don't understand why int fahr = celsius / 5 * 9 + 32; works but not int fahr = 9/5 * celsius+32;

For the former, you probably declared celsius as a float or as a double, making the entire right side evaluate as such. Assuming you used float, it works out like this:

celsius / 5 * 9 + 32
(celsius / 5.0) * 9 + 32
(celsius / 5.0 * 9.0) + 32
(celsius / 5.0 * 9.0 + 32.0)

For the latter, 9/5 is integer arithmetic that evaluates to 1 before the rest of the math happens as floating point. In this case:

9 / 5 * celsius + 32
1 * celsius + 32 // Because of this, you get an incorrect answer
celsius + 32
celsius + 32.0

Note the the type of the left hand side is irrelevant; the right-hand side is evaluated without regard to that.

Update: You said celsius is an int, which means you just happened to get lucky and test with a value that is a multiple of 5, giving you a correct integer result to celsius / 5 before doing valid integer arithmetic for the rest of the statement. In your second example, being a multiple of 5 doesn't help you.

In any case, now you know why you got lucky, but the linked question gives you the answer to what you actually need to to do have a formula that works when celsius isn't a multiple of 5: use floating point math, as demonstrated in all of the answers there.