Division by 0.0 allowed

[Dr. Dri]

with this code i get an error at run time

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l.l = 0
m.l = 1 / l

but not with this one

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f.f = 0.0
g.f = 1.0 / f

but still i can't do that because the compiler says that division by zero is forbidden even for floats

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g.f = 1.0 / 0.0

is it possible to remove this check ?

Dri

[michaeled314]

You're crazy if you think you will get a result for a divide by zero!

[Trond]

1/0 is undefined. So allowing it makes no sense.

Edit: Rrrrrr too late for the second time today.

[Guimauve]

Divide by zero ...

It's a joke or ...

The divide by zero without error message it's a MAJOR bug. The best way to built your code is this case is :

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Alpha.f = 0.0

If Alpha <> 0.0
   Gamma.f = 1.0 / Alpha
EndIf

Regards
Guimauve

[Kaeru Gaman]

but the result holds the correct return-value of the FPU:

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f.f = 0.0
g.f = 1.0 / f
Debug g

debugs

1.#INF

[Derek]

is it possible to remove this check ?

I have to ask, why would you want to?

[Deeem2031]

The divide by zero without error message it's a MAJOR bug.

Its no bug, but if u really want to have an exception when divide by zero with float-numbers try this:

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FPU_ControlWord.w
!FSTCW [v_FPU_ControlWord]
FPU_ControlWord ! 4
!FLDCW [v_FPU_ControlWord]

So you get your "error message":

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zero.f = 0.0

Debug 1.0 / zero

FPU_ControlWord.w
!FSTCW [v_FPU_ControlWord]
FPU_ControlWord ! 4
!FLDCW [v_FPU_ControlWord]

Debug 1.0 / zero

[Trond]

The divide by zero without error message it's a MAJOR bug.

It's not really major in my opinion, but I reported it a few days ago anyways.

but the result holds the correct return-value of the FPU:

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f.f = 0.0
g.f = 1.0 / f
Debug g

debugs

1.#INF

There is no "correct" return value. 1/0 is not infinity.

Division by zero is an operation for which you cannot find an answer, so it is disallowed. You can understand why if you think about how division and multiplication are related.

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   12 divided by 6 is 2   because
    6 times 2 is 12

   12 divided by 0 is x   would mean that
    0 times x = 12

But no value would work for x because 0 times any number is 0. So division by zero doesn't work.

[Deeem2031]

There is no "correct" return value. 1/0 is not infinity.

We are not talking about what u've learned in school - you are writting programs which uses the FPU, and if you ask the FPU 1/0 is infinity! (or an exception if you set the ZeroFlag of the FPU ControlWord (see code above))

[Trond]

There is no "correct" return value. 1/0 is not infinity.

We are not talking about what u've learned in school - you are writting programs which uses the FPU, and if you ask the FPU 1/0 is infinity! (or an exception if you set the ZeroFlag of the FPU ControlWord (see code above))

You are writing programs in PB and if you ask PB it's not allowed.

[Kaeru Gaman]

PB does not allow it hardcoded, thats why

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g.f = 1.0 / 0.0

fails: the compiler doesn't accept the expression.
sure he couldn't because he tries to evaluate constant expressions.

but when you use variables, PB passes the responsability to the FPU,
that will react according to it's ControlWord, like Deeem described.

to me this behaviour is absolutely correct.
I only wish we had a native command to set the FPU ControlWord,
and a check to compare a result with 1.#INF

[thefool]

//some theory based on my actual thinking//

For sure a/0 is NOT infinite, thats bullshit. The problem is that floats are rather unprecise, and 0/VERYSMALLNUMBER = a very very large number.

So if the FPU finds a/0 to be infinite its because its not precise.

We are not talking about what u've learned in school

We certainly are! All of this is about that. Now i don't know what your education is but this is actually very very low level math..

[Trond]

PB does not allow it hardcoded, thats why

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g.f = 1.0 / 0.0

fails: the compiler doesn't accept the expression.
sure he couldn't because he tries to evaluate constant expressions.

but when you use variables, PB passes the responsability to the FPU,
that will react according to it's ControlWord, like Deeem described.

Not true. PB uses doubles and quads for evaluating constant expressions.

See how there's a rounding error? That's because a double was used.

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a.q = 23372036854775807.0
Debug a

Edit: No, it's not a double, I think it's a ten-byte float. But it's a float rounding error.

[Kaeru Gaman]

it seems to be a full FPU register. thats cool to know.
so if I use a decimal point in a literal, it will be evaluated by the FPU during the compile.

[DarkDragon]

http://en.wikipedia.org/wiki/Infinity#In_computing
http://en.wikipedia.org/wiki/NaN