C Language: Initializing float variable with a calculation

Question

I haven't done a lot of floating point math programming in any language let alone in C.

I'm writing a temperature conversion program as an exercise and have a question about floating point numbers. I have a code frag as listed below. In both cases Temp1 and Temp2 are 0.0 when P_FahrenheitTemp is <> 32.0. However, if I use the CF3 factor in the calculation the LOC VERRKKKS!!! :-)

This seems intuitively obvious to me but... Is this compiler dependent or is a cast operator necessary on the initialization? BTW, I'm writing this code on an IBM iSeries platform using the C/C++ compiler which strictly adheres to ASNI and ISO standards.

Thank you in advance for any info!

Martin Kuester

#define CF3 5/9;

float Conv2Celsius(float P_FahrenheitTemp)
{
  float Temp1, Temp2, Temp3;
  float ConvAdj = 32.0;
  float CF1 = 0.555556;
  float CF2 = 5/9;

  //[°C] = ([°F] - 32) × 5/9
  Temp1 = (P_FahrenheitTemp - ConvAdj) * CF1; 
  Temp2 = (P_FahrenheitTemp - ConvAdj) * CF2;
  Temp3 = (P_FahrenheitTemp - ConvAdj) * CF3;

  return(Temperature);
}

WRONG: `#define CF3 5/9;` Better: `#define CF3 5.0/9.0`. 1) Use floating point constants (else you risk [integer truncation](https://en.wikipedia.org/wiki/Truncation). 2) #define macros typically should NOT have a ";" terminator. — paulsm4, Oct 25 '21 at 19:22
`5/7` is an intereger division, so the result is truncated to an integer. It works in a macro because `(...) * 5 / 7` that you get after expansion is parsed as `((...) * 5) / 7`. — HolyBlackCat, Oct 25 '21 at 19:22
what is `return(Temperature);`? There is no `Temperature` variable in sight. — bolov, Oct 25 '21 at 19:23
@HolyBlackCat Or `inline constexpr double CF3 = 5.0/9.0;` because why is a macro necessary for this at all — Nathan Pierson, Oct 25 '21 at 19:29
@NathanPierson or `constexpr double CF3 = 5.0/9.0;` because `constexpr` implies `inline` — bolov, Oct 25 '21 at 19:30
or `constexpr double CF3 = 5.0 / 9.0;` because I can't stand no spaces around binary operators. We can do this aaaaall day. — bolov, Oct 25 '21 at 19:31
@bolov I believe that only applies to data members of classes, not file scope variables. — Nathan Pierson, Oct 25 '21 at 19:31
@463035818_is_not_a_number *what is "LOC VERRKKKS"* A lake in Scotland? — user4581301, Oct 25 '21 at 20:00
@ 463035818_is_not_a_number, user4581301 likely "the LOC VERRKKKS!!!" --> "the line-of-code (surprisingly) works!". — chux - Reinstate Monica, Oct 25 '21 at 21:58
Thank you all for your comments and explanations. Some clarifications: 1. LOC = Line of Code 2. VERRRKKKKSSS! Is an obscure Germanic word meaning "IT VERRRRRKKKKKKKKSSSSS!" 3. C++ is rooted in C Language. My question re: floating point variable initialization would apply to either language. I figured tagging C++ would increase the number of readers. Conclusions: 1. Initializing float variables use 'f' in the literal. 2. Cast as float for literal calculation so that result is explicitly float. Martin — Martin Kuester, Oct 26 '21 at 12:06

chux - Reinstate Monica · Answer 1 · 2021-10-25T22:00:31.650

Let us look closer.

  float CF1 = 0.555556;
  Temp1 = (P_FahrenheitTemp - ConvAdj) * CF1; 
  // same as 
  Temp1 = (P_FahrenheitTemp - ConvAdj) * (float) 0.555556;

  float CF2 = 5/9;
  Temp2 = (P_FahrenheitTemp - ConvAdj) * CF2;
  // same as 
  float CF2 = 0;  // 5/9 is integer division
  Temp2 = (P_FahrenheitTemp - ConvAdj) * 0;

  #define CF3 5/9
  Temp3 = (P_FahrenheitTemp - ConvAdj) * CF3;
  // same as 
  Temp3 = (P_FahrenheitTemp - ConvAdj) * 5 / 9;
  Temp3 = (P_FahrenheitTemp - ConvAdj) * 5.0f / 9;
  //      ^--- float  multiplication -------^             
  // same as 
  Temp3 = (P_FahrenheitTemp - ConvAdj) * 5.0f / 9.0f;
  //      ^--- float divsion ----------------------^

Temp3 "VERRKKKS" because it is not scaling by 5/9. Instead it is a text substitution in the line-of-code and so multiplies by 5 and then divides by 9.

Temp3 is correct and best of the three.
Temp1 is almost correct as not as certainly precise * (float) 0.555556 as * 5.0f/9.0f.
Temp2 is wrong as the answer is always 0, even when it should not be

I have a code frag as listed below. In both cases Temp1 and Temp2 are 0.0 when P_FahrenheitTemp is <> 32.0.

Temp1 is not 0.0.

To set aside the minor additional error in the constant, use at least 9 digits with float and a f suffix.

//float CF1 = 0.555556;
float CF1 = 0.555555556f;

Suggested replacement

float Conv2Celsius(float P_FahrenheitTemp) {
  float ConvAdj = 32.0f;
  float CF = 5.0f/9.0f; // or 0.555555556f

  //[°C] = ([°F] - 32) × 5/9
  return (P_FahrenheitTemp - ConvAdj) * CF; 
}

I knew I was missing something -- I just couldn't see what. Thank you. — David C. Rankin, Oct 25 '21 at 19:56

score 1 · Answer 2 · answered Oct 25 '21 at 19:54

In C language the line float CF2 = 5/9; will be processed as follows:

The right side of the assignment operator 5/9 is evaluated first. The compiler here sees two integer values divided on each other so it will save the result into an integer temporary variable. This will lead to truncation of the fractional part of the actual result 0.555556 to 0.
The result will be assigned then to CF2.

What to do?

Alot of options; float CF2 = 5.0/9; or float CF2 = (float)5/9; or even float CF2 = 5./9;

the same with CF3

C Language: Initializing float variable with a calculation

2 Answers2