List of strings

SOFiSTiK Teddy
#1

I’ve been trying to create a list of strings to name load cases in a loop.

However, it would seem that it’s not possible to interact with a list of strings after it’s created, is that really true?

Please see the example code piece below:

+prog template
    head List of strings

    let#list1 1,2,3
    prt#list1

    let#list2 'A','B','C'
    prt#list2

    let#list3 #list1(:)
    prt#list3

    let#list4 #list2(:)
    prt#list3
end
#2

Hello,

If you want to increase the title of an load case set, try to use the following code snippet. The first short code creates a load case with the title “2C”.

LET#list1 'A','B','C'
LET#list2 '1','2','3'
LC 1 DLZ 1.0 TITL #list2(1)#list1(2)

Or you work with an increasing variable name:

LET#lf 2
LET#s1 "loadcase #(lf,3)"
PRT#s1

Best regards
Frederik Höller
Your SOFiSTiK Support Team

#3

thank you for your reply. Even though you have now edited your reply, I hope you will still let the responsible developer know about this issue.

Meanwhile, I’ll do the workaround.

#4

Hello,

I have informed the responsible developer but due to the lack of necessity, the program behaviour is not changed.

Best regards
Frederik Höller
Your SOFiSTiK Support Team

#5

Okay, I’d say that debugging your software is rather necessary, but I suppose priorities do vary.
Furthermore I’d suggest increasing the allowable string length to more than 8 characters as well.

But I consider the thread closed for now.

Thank you.

#6

While we are at the topic of Let-variables and their implementation:
There seems to be a difference in how whitespace is treated whether the type is a string or a numeric value.

The assignment to #list3 has a bug in it (inserting blanks in the vector).
Also: Are you supposed to be able to swap commas for whitespace?

Example input:

let#list1 ‘A’,‘B’,‘C’
let#list2 1,2,3
prt#list1
prt#list2

let#list3 ‘A’, ‘B’, ‘C’
let#list4 1, 2, 3
prt#list3
prt#list4

let#list5 ‘A’ ‘B’ ‘C’
let#list6 1 2 3
prt#list5
prt#list6

Corresponding output:

LET#LIST1 ‘A’,‘B’,‘C’
LET#LIST2 1,2,3
PRT#LIST1
CADINT VARIABLE LIST1 ( 0) = "A "
CADINT VARIABLE LIST1 ( 1) = "B "
CADINT VARIABLE LIST1 ( 2) = "C "
PRT#LIST2
CADINT VARIABLE LIST2 ( 0) = 1.000
CADINT VARIABLE LIST2 ( 1) = 2.000
CADINT VARIABLE LIST2 ( 2) = 3.000

LET#LIST3 ‘A’, ‘B’, ‘C’
LET#LIST4 1, 2, 3
PRT#LIST3
CADINT VARIABLE LIST3 ( 0) = "A "
CADINT VARIABLE LIST3 ( 1) = " "
CADINT VARIABLE LIST3 ( 2) = "B "
CADINT VARIABLE LIST3 ( 3) = " "
CADINT VARIABLE LIST3 ( 4) = "C "
PRT#LIST4
CADINT VARIABLE LIST4 ( 0) = 1.000
CADINT VARIABLE LIST4 ( 1) = 2.000
CADINT VARIABLE LIST4 ( 2) = 3.000

LET#LIST5 ‘A’ ‘B’ ‘C’
LET#LIST6 1 2 3
PRT#LIST5
CADINT VARIABLE LIST5 ( 0) = "A "
CADINT VARIABLE LIST5 ( 1) = "B "
CADINT VARIABLE LIST5 ( 2) = "C "
PRT#LIST6
CADINT VARIABLE LIST6 ( 0) = 1.000
CADINT VARIABLE LIST6 ( 1) = 2.000
CADINT VARIABLE LIST6 ( 2) = 3.000

#7

I appreciate that you both are testing the software so thoroughly.
But in this case you are expecting functionalities that are not programmed for this use case.
As far as I know, string arrays should not be used. Just use one combination of literals for one Variable.
For example

LET#TEXT ‘Hello world’
and not
LET#TEXT ‘Hello’,’ world’

The behaviour of the string array in your short example is interesting, but I can’t explain why you get these results. Just stay away from string arrays :wink: