string slicing using len() - Printable Version

string slicing using len() - Printable Version

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string slicing using len() - programandala.net - 04-19-2010

I didn't understand why a piece of my program didn't work as expected. I suspected something was wrong with the for-next loop. I tried some tests, simplified versions of the problem, and finally find out the following:

Code:
dim text as string

dim i as ubyte

let text="hello world!"

cls

print "Fine:"

for i = 0 to len(text)-1

    print i,text(to i)

next i

pause 0

cls

print "All texts are complete!:"

for i = 0 to len(text)-1

    print i,;

    let text = text(to len(text)-1)

    print text

next i

stop

It was a surprise. The second loop prints the same original text every time.

I tried the equivalent code in Sinclair Basic, and of course it worked as expected:

Code:
LET A$="hello world!"

FOR I=0 TO LEN(A$)-1

PRINT I,;

LET A$=A$(TO LEN(A$)-1)

PRINT A$

NEXT I

STOP

Then I tried it without the loop:

Code:
dim text as string

let text="hello world!"

cls

print text

let text = text(to len(text)-1)

print text

let text = text(to len(text)-1)

print text

let text = text(to len(text)-1)

print text

let text = text(to len(text)-1)

print text

let text = text(to len(text)-1)

print text

let text = text(to len(text)-1)

print text

let text = text(to len(text)-1)

print text

stop

And the problem remained.
Then I hardcoded some slicing values:

Code:
dim text as string

let text="hello world!"

print text

let text = text(to 10)

print text

let text = text(to 9)

print text

let text = text(to 8)

print text

let text = text(to 7)

print text

let text = text(to 6)

print text

let text = text(to 5)

print text

stop

And it worked fine.

Is there something I'm missing in my code or is it a compiler issue?

Re: string slicing using len() - boriel - 04-19-2010

It happens string slicing starts from 0 (not from 1). So it ranges from 0 to LEN(a$)-1.
If you want your program work OK, you should do:

Code:
let a$="hello world"

let l=len(a$)-1

for i = 1 to l:

    let a$=a$(to len(a$) - 2)

    print i, a$

next i

Notice also that FOR upper limit is evaluated ON EACH iteration (like C). So if len(a$) changes, the loop will shorten. So better use a temporary var l to store initial LEN(a$).

Arrays subscripts starts from 0 to N-1 (like in C). But this behavior can be changed using --sinclair or --array-base switch.

NOTE: I'm planning (for compatibility) an --string-base (so LEN and string-slicing will work exactly as in Sinclair BASIC, by specifying --string-base=1). Also with FOR, stating --constant-for etc...

Re: string slicing using len() - programandala.net - 04-19-2010

boriel Wrote:Notice also that FOR upper limit is evaluated ON EACH iteration (like C). So if len(a$) changes, the loop will shorten. So better use a temporary var l to store initial LEN(a$).

I learned it while debugging. It's an important difference with Sincllar Basic.

boriel Wrote:Arrays subscripts starts from 0 to N-1 (like in C). But this behavior can be changed using --sinclair or --array-base switch.

I forgot it! Thanks. One of my tests used "-2" intead of "-1" because of that. Anyway, the problem arises when trying to remove the last char of the string. Then "-2" cannot be used. I tried the following expression:

Code:
let text=text( to len(text)-2+(len(text)=1))

It doesn't work because

Code:
(len(text)=1)

doesn't return "1" when true, as I expected (Sinclair Basic does).

I opened a new thread about this different issue.

Re: string slicing using len() - boriel - 05-03-2010

A new compiler version (r1564) has been uploaded. It allows you to use --string-base parameter to tell the compiler which is the 1st base-index position in strings:

Code:
./zxb.py --string-base=1 ...

The same applies for array (--array-base). So if you want arrays to start at index 1, use --array-base=1.
Also using --sinclair enables all this compatibilities with a single parameter.