simple if..else struc problem

[Mr. Carlos Repolho 1]

if (Lratio < 1)

 

Lfn.setvalue(Los)

 

elseif (Lratio >= 1.1)

 

Lfn.setvalue(1.0667*L)

 

else

 

Lfn.setvalue(L + (2/3)*(Los - L))

 

endif

 

For some reason for a Lratio =1.05, the code jumps the else command and attributes zero to Lfn.

 

Also if i would like to ask in another case if i would want to write something like this:

 

elseif( 1=<Lratio<1.1) .. could it be done like: elseif ( and ( Lratio>=1,Lration<1.1))

 

best regards 

 

 

 

[Mr. Claus Abt 14]

HI Carlos,

the jump to else seems reasonable since 1.05 is neither smaller than 1 nor larger than 1.1. It is a bit difficult to understand why Lfn comes out as zero if exept for Los= -0.5 x L

Cheers

Claus

[Stefan Wunderlich 6]

Hi Carlos,

 

the "and(..., ...)" should be what you are looking for.

 

Cheers,

Stefan

[Mr. Carlos Repolho 1]

Good morning 

 

I also dont understant why it jumps the else Claus.

 

Stefan can you give me a simple example how to use the and structure in this case, i already tried it before but it didnt work.

 

best regards

[Stefan Wunderlich 6]

Hi Carlos,

 

what do you mean by "jumps the else", does it run into the else case if Lratio == 1.05?

The "and" statement works like this:

 

if (Lratio < 1)

 

Lfn.setvalue(Los)

 

elseif (Lratio >= 1.1)

 

Lfn.setvalue(1.0667*L)

 

elseif ( and ( Lratio>=1, Lratio<1.1))

 

Lfn.setvalue(L + (2/3)*(Los - L))

 

endif

 

But that does not make any sense since "and ( Lratio>=1,Lratio<1.1)" is implicit for the else case in your example.

 

Cheers,

Stefan

[Mr. Carlos Repolho 1]

"But that does not make any sense since "and ( Lratio>=1,Lratio<1.1)" is implicit for the else case in your example"

 

Yes i thought it would be implicit due to the value 1.05 complies with the inner interval, but for some reason it jumps of the 'else '  and doesn't attribute any value to Lfn.

 

Anyway i will try the and command and then after i will  upload the file so you could take a look if you have some spare time.

 

This is a feature for a simple empirical method of resistance prediction called Hollenbach, in the beginning of the code i need to define this Lfn which depends on a ratio that is obtained by (Los/lpp), after the Fn is calculated and the prediction is made.

 

After this code i still need to do Holtrop&Mennen and the EEDI so i was really needing some help in this :)

 

best regards

 

Thanks again!

[Stefan Wunderlich 6]

For debugging purpose, you could add some "echo" commands, to see which branch it really takes (you will see the output in console):

 

if (Lratio < 1)

    echo("Lratio < 1: " + Lratio)

    Lfn.setvalue(Los)

elseif (Lratio >= 1.1)

    echo("Lratio >= 1.1: " + Lratio)

    Lfn.setvalue(1.0667*L)

else

    echo("else branch: " + Lratio)

    Lfn.setvalue(L + (2/3)*(Los - L))

endif

 

Cheers,

Stefan

[Mr. Carlos Repolho 1]

I saw the tutorial and i thought about doing it, will do it as soon as i leave work. doing my thesis while working at a shipyard has become a much bigger challenge than i was prospecting... :wacko:

 

When i have some results i will post here.

 

Danke Stefan

[Mr. Carlos Repolho 1]

Hi !

 

Something strange its happening! The code works and it worked with what i had written before but when debugging if i don't put the break point exactly on the else, Lfn variable it jumps and for the porpuse of debugging the user cant see any change.

 

I think is something about the debugger so if i could send the feature to you i think it would be usefull.

 

best regards and thanks