Basic math issue in C using cygwin

Question

I am not sure what's going on here, but I have a code sample which is not producing what I expect, here is part in question (just a section):

printf("Now enter a Fahrenheit to convert to Centigrade\n");
fflush(stdout);
scanf("%f",&c);
d = (c-32)*(5/9);
printf("%f\n",c);
printf("Your result in Centigrade is %f\n",d);

The result when I enter 212 for the input is: Your result in Centigrade is 0.000000

When I make a small change:

printf("Now enter a Fahrenheit to convert to Centigrade\n");
fflush(stdout);
scanf("%f",&c);
d = (c-32)*5/9;
printf("%f\n",c);
printf("Your result in Centigrade is %f\n",d);

The result now when I enter 212 for the input is: Your result in Centigrade is 100.00000, which is correct.

In math (a+b) * (c/d) is the same as (a+b) * c / d; so what's going on with the zero result with just the extra parentheses?

/*Sorry this does not work in Java, it suffers the same issue as with C.*/ float c = ((a-32)* 5/9); //this works — Christopher Keveny, Jul 10 '16 at 01:16
Possible duplicate of [What is the behavior of integer division in C?](http://stackoverflow.com/questions/3602827/what-is-the-behavior-of-integer-division-in-c) — Mark Rotteveel, Jul 10 '16 at 10:07

score 3 · Answer 1 · answered Jul 10 '16 at 00:47

3

Since both operands are integers, 5/9 results in 0 since integer division is performed. Use 5./9 instead.

answered Jul 10 '16 at 00:47

Ignacio Vazquez-Abrams

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Okay, so in C undefined operands bound by parentheses will default to the higher precedent data type? – Christopher Keveny Jul 10 '16 at 00:56
@ChristopherKeveny not exactly. "the usual arithmetic conversions" apply, and then possibly integer promotions. See http://c0x.coding-guidelines.com/6.3.1.8.html or even better -- the c standard itself. – Paul Hankin Jul 10 '16 at 01:01
Not just "by parentheses", but always. – Ignacio Vazquez-Abrams Jul 10 '16 at 01:01
Thank you everyone. – Christopher Keveny Jul 10 '16 at 01:31
Be aware that `5.` is a `double` value; `5.0f` or `5.0F` is a `float` value. Since `c` is a `float`, there will be conversions from `float` to `double`, and if `d` is a `float` too, then there will be conversions back from `double` to `float`. – Jonathan Leffler Jul 10 '16 at 01:57

sps · Answer 2 · 2016-07-10T01:13:00.867

Here, you need to be careful about two things.

Operator preceedence
Automatic type conversion

First, () operator has higher preceedence than * or / operators.

Second , if an expression contains a float and and int, the int will be automatically converted to float.

Now in your first case what happens is:

(c - 32) is calculated, (here 32 is converted to 32.0 and calculation is done). (say this equals x)
(5/9) is calculated. Here both expressions are int so nothing is converted to float. And for int calculation, 5/9 equals to 0. Say this equals y
Your final answer is calculated as x * y, and since y is 0 , your answer is 0.0.

However in your second case what happens is:

(c - 32) is calculated, (here 32 is converted to 32.0 and calculation is done).
(c - 32) * 5 is calculated. Here 5 is converted to 5.0 and calculation is done. (say this equals x : x is float)
x / 9 is calculated. Here 9 is converted to 9.0 and calculation is done.

So, you can see in second case, there was not multiplication with a 0 - like in the first case - hence you are getting a non-zero (hopefully correct) answer.

Thank you sps. The automatic background casting of the fractional quotient to int caught me. (int)(5/9) indeed is zero. — Christopher Keveny, Jul 10 '16 at 01:30
@ChristopherKeveny: `(5/9)` is zero even without the cast; both operands are integers, so the calculation is done using integer arithmetic. — Jonathan Leffler, Jul 10 '16 at 01:55

Basic math issue in C using cygwin

2 Answers2