Float problem

**+Azazel+** · 02-27-2008

Hey guys! I think I must be going crazy...

Code:

int main()
{
  float test = 48 / 512;
  printf("%f", test);
  return (0);
}

gives me 0.0000. Why? How can I get the answer?

**Banana Man** · 02-27-2008

48 / 512 is less than 0. Both 48 and 512 are integers so the whole division is done on integers and the precision is lopped off leaving you with 0. Change one or both of the numbers to float and it'll work.

Code:

int main()
{
  float test = 48.0f / 512.0f;
  printf("%f", test);
  return (0);
}

**esbo** · 02-27-2008

Originally Posted by Banana Man

48 / 512 is less than 0. Both 48 and 512 are integers so the whole division is done on integers and the precision is lopped off leaving you with 0. Change one or both of the numbers to float and it'll work.

Code:

int main()
{
  float test = 48.0f / 512.0f;
  printf("%f", test);
  return (0);
}

You don't need to add the f at the end of the value, a decimal point is enough to make
it work. Infact if you miss out the decimal point it won't compile on my machine!!

Code:


int main()
{
 float test = 48f/512f;
  printf("%f", test);
  return (0);
}

Code:

invalid sufix 'f' on integer constant,

So I am not sure what the f does,

**robwhit** · 02-27-2008

1.0 is a double. 1.0f is a float.

**esbo** · 02-27-2008

Originally Posted by robwhit

1.0 is a double. 1.0f is a float.

The result is a float though - well it works anyway, so what difference does it make?

**Elysia** · 02-27-2008

Originally Posted by esbo

You don't need to add the f at the end of the value, a decimal point is enough to make
it work. Infact if you miss out the decimal point it won't compile on my machine!!

Leaving it out will do integer division, hence giving you 0. That is no error or warning.
If you add decimals, you get a double, but some compilers won't warn about that.

So I am not sure what the f does,

So again, as others stated, it makes the number a float. It needs to be 1.0f, not 1f.

And for once, you used "int main" and "return 0". Congratulations! Must have been hard on you.

**esbo** · 02-27-2008

Originally Posted by Elysia

Leaving it out will do integer division, hence giving you 0. That is no error or warning.

Not if you add a decimal point it won't.

Originally Posted by Elysia

If you add decimals, you get a double, but some compilers won't warn about that.
So again, as others stated, it makes the number a float. It needs to be 1.0f, not 1f.

It only make it a float if it has a decimal point, otherwise it just gives you a compiler error.

Originally Posted by Elysia

And for once, you used "int main" and "return 0". Congratulations! Must have been hard on you.

Not really, I am quite good at cut and paste.

**Elysia** · 02-27-2008

Originally Posted by esbo

Not if you add a decimal point it won't.

But that was beyond the point. The point was that if you didn't do it, it would be integer division.

It only make it a float if it has a decimal point, otherwise it just gives you a compiler error.

Of course. A float is a floating point number, after all.

Not really, I am quite good at cut and paste.

Yes, but you could have removed them, to better suit your "tastes."

**esbo** · 02-27-2008

Originally Posted by Elysia

But that was beyond the point. The point was that if you didn't do it, it would be integer division.

Of course. A float is a floating point number, after all.

Yes, but you could have removed them, to better suit your "tastes."

The f changes a floating point number to a float, it won't change an integer to a float.
A double is a floating point number.
My 'taste' is just to copy what is already there, I won't change something if it compiles
and works.
I don't really see the point in having an f there at all, it seems rather pointless.

**VirtualAce** · 02-27-2008

The f changes a floating point number to a float, it won't change an integer to a float.
A double is a floating point number.
My 'taste' is just to copy what is already there, I won't change something if it compiles
and works.
I don't really see the point in having an f there at all, it seems rather pointless.

Because it is a very bad habit to get into. If you wish to use float then specify the f otherwise do not specify the f. It also informs other programmers that you are using float's and not doubles. There is a big difference between a double and a float. Just because 'it works' does not mean it's correct.

This:

Code:

float x = 4f;

This means nothing to me except that the person who programmed it either was very tired that day or doesn't understand floating point values. Did they want 4.0f or is it a typo?

This:

Code:

float x = 4.0f;
double y = 4.0;

Means that x is a float and y is a double.

So you ask why use the f? Here is an example:

Code:

float x = some_angle * 180.0 / 3.14159f

What do you suppose happens in this assignment?

**esbo** · 02-27-2008

Originally Posted by Bubba

Because it is a very bad habit to get into. If you wish to use float then specify the f otherwise do not specify the f. It also informs other programmers that you are using float's and not doubles. There is a big difference between a double and a float. Just because 'it works' does not mean it's correct.

This:

Code:

float x = 4f;

This means nothing to me except that the person who programmed it either was very tired that day or doesn't understand floating point values. Did they want 4.0f or is it a typo?

This:

Code:

float x = 4.0f;
double y = 4.0;

Means that x is a float and y is a double.

So you ask why use the f? Here is an example:

Code:

float x = some_angle * 180.0 / 3.14159f

What do you suppose happens in this assignment?

I don't care, I would not have used the f so it's not really my problem.
What do you think happens?

Also

Code:

float x = 4.0;
double y = 4.0;

Means that x is a float and y is a double.

**whiteflags** · 02-27-2008

>> Means that x is a float and y is a double.

Maybe, but it's an issue in precision: in Bubba's example, we're using a floating-point constant. Had he left off the trailing f, we would have done double floating point division, and the answer would have been rounded to fit in a float variable.

As you should know, the quality in answers is quite different when finer precision matters. So, it's at least fairly important to avoid conversion issues while you're using constants like pi, or some other number.

**esbo** · 02-27-2008

Originally Posted by citizen

>> Means that x is a float and y is a double.

Maybe, but it's an issue in precision: in Bubba's example, we're using a floating-point constant. Had he left off the trailing f, we would have done double floating point division, and the answer would have been rounded to fit in a float variable.

And exactly what is wrong with that?

Originally Posted by citizen

As you should know, the quality in answers is quite different when finer precision matters. So, it's at least fairly important to avoid conversion issues while you're using constants like pi, or some other number.

I am not really sure what you are saying here, could you be more specific about what point
it is your are trying to make?

**esbo** · 02-27-2008

Also I have been getting into bad habits doing this.
In 'real life' I would never write:-

Code:

float x = 4.0;
double y = 4.0;

I would write:-

Code:

float x = 4;
double y = 4;

Means that x is a float, y is a double

And never in a million years (I hope) would I write

Code:

float x = 4.0f;
double y = 4.0;

**whiteflags** · 02-27-2008

float x = 4;
double y = 4;

Nothing wrong with promotion (as long as it is compatible). But when you convert to a lesser data type, you might lose your data.

>> And exactly what is wrong with [rounding]?
>> what point it is your are trying to make?

It is completely reasonable to avoid possible loss of data.

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