CODEWITHSUNDEEP
ctypes
Thomas Owens
Thomas Owens

Reputation: 116161

Why does this C code produce a double instead of a float?

celsius = (5.0/9.0) * (fahr-32.0);

Is it just a development choice that the C developers decided upon or is there a reason to this? I believe a float is smaller than a double, so it might be to prevent overflows caused by not knowing what decimal format to use. Is that the reason, or am I overlooking something?

Upvotes: 5

Views: 937

Answers (5)

amaterasu
amaterasu

Reputation: 1080

Back in the day of K&Rv1, it was encouraged to use float/double interchangeably being as all expressions with floating-point types were always evaluated using `double' representation, a problem in cases where efficency is paramount. A floating-point constant without an f, F, l, or L suffixed is of type double. And, if the letter f or F is the suffix, the constant is of type float. And if suffixed by the letter l or L, it is of type long double.

Upvotes: 1

hoyhoy
hoyhoy

Reputation: 6351

The reason that the expression is cast to double-precision is because the literals specified are double-precision values by default. If you specify the literals used in the equation as floats, the expression will return a float. Consider the following code (Mac OS X using gcc 4.01). 

#include <stdio.h>
int main() {
  float celsius;
  float fahr = 212;
  printf("sizeof(celsius) ---------------------> %d\n", sizeof(celsius));
  printf("sizeof(fahr) ------------------------> %d\n", sizeof(fahr));
  printf("sizeof(double) ----------------------> %d\n", sizeof(double));
  celsius = (5.0f/9.0f) * (fahr-32.0f);
  printf("sizeof((5.0f/9.0f) * (fahr-32.0f)) --> %d\n", sizeof((5.0f/9.0f) * (fahr-32.0f)));
  printf("sizeof((5.0/9.0) * (fahr-32.0)) -----> %d\n", sizeof((5.0/9.0) * (fahr-32.0)));
  printf("celsius -----------------------------> %f\n", celsius);
}

Output is:

sizeof(celsius) ---------------------> 4
sizeof(fahr) ------------------------> 4
sizeof(double) ----------------------> 8
sizeof((5.0f/9.0f) * (fahr-32.0f)) --> 4
sizeof((5.0/9.0) * (fahr-32.0)) -----> 8
celsius -----------------------------> 100.000008

Upvotes: 3

Shog9
Shog9

Reputation: 159590

celsius = (5.0/9.0) * (fahr-32.0);

In this expression, 5.0, 9.0, and 32.0 are doubles. That's the default type for a floating-point constant - if you wanted them to be floats, then you would use the F suffix:

celsius = (5.0F/9.0F) * (fahr-32.0F);

Note that if fahr was a double, then the result of this last expression would still be a double: as Vaibhav noted, types are promoted in such a way as to avoid potentially losing precision.

Upvotes: 27

dmckee --- ex-moderator kitten
dmckee --- ex-moderator kitten

Reputation: 101151

Floating point constants should have the available highest precision. The result can be assigned to a float without undue trouble.

Upvotes: 0

Vaibhav
Vaibhav

Reputation: 11436

I think the reason is to ensure that any result can be encompassed. so the natural choice is double as it is the largest data type.

Upvotes: 3

Related Questions