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Bug or feature?

Slawa

 Hello

 In my script the following is used
x1=y/0.01
x1t=truncate(x1)
If x1 - an integer that x1t should be equal in a case x1 but

if y=1308.1 then
x1=130810.0
x1t=130809

Why??

Parents

  • Hi,

    Neither a bug or a feature!  You show x1 as being 130810.0, but it may actually be internally represented as 130809.999997 (say), so it may display as 130810.0, but when you truncate the fractional part to make it an integer, you see 130809.  You can print more digits of a floating point number, e.g. using printf() to check this for yourself, something like the following:

     printf("%6.8f" x1) 
     => 130809.99999700

    The default print representation of a floating point may give you up to 6 digits of precision, but only for about 7 significant digits, so if the number is already 6 digits 'wide', then you will only get one digit of precision and the number is rounded when it is presented to you.

    I hope that this helps to answer your question, or at least helps you to understand what is happening.

    Regards,

    Lawrence.

Reply

  • Hi,

    Neither a bug or a feature!  You show x1 as being 130810.0, but it may actually be internally represented as 130809.999997 (say), so it may display as 130810.0, but when you truncate the fractional part to make it an integer, you see 130809.  You can print more digits of a floating point number, e.g. using printf() to check this for yourself, something like the following:

     printf("%6.8f" x1) 
     => 130809.99999700

    The default print representation of a floating point may give you up to 6 digits of precision, but only for about 7 significant digits, so if the number is already 6 digits 'wide', then you will only get one digit of precision and the number is rounded when it is presented to you.

    I hope that this helps to answer your question, or at least helps you to understand what is happening.

    Regards,

    Lawrence.

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