Dump overflow convt_overflow in a function

[Former Member]

Hello master.

I created a function that calculate a number from a string of numbers.

I checked and it works, but the string that needs to use the function is long

cadena = 0200000005500002133200000154700076945190

cadena that i need to work with is 40 lenght. I checked it with 30 lenght and it worked. how can i make it work with a lenght of 40???

FUNCTION zcalculo_dv.
*"----------------------------------------------------------------------
*"*"Interfase local
*"  IMPORTING
*"     REFERENCE(CADENA) TYPE  STRING
*"  EXPORTING
*"     REFERENCE(DV) TYPE  C
*"----------------------------------------------------------------------
* Toma una cadena de numeros, solo numeros (sin .,"·%!$) y calcula el DV

  DATA: suma TYPE i,
        mult TYPE i,
        num  TYPE string.

  num = cadena.

  mult = 2.
  suma = 0.

  WHILE num <> 0.
    suma = suma + num MOD 10 * mult.
    num = num DIV 10.
    mult = mult + 1.
    IF mult > 7.
      mult = 2.
    ENDIF.
  ENDWHILE.

  num = 11 - suma MOD 11.
  IF num = 10.
    dv ='K'.
  ELSEIF num = 11.
    dv = '0'.
  ELSE.
    MOVE num TO dv.
  ENDIF.

ENDFUNCTION.

It should return a number. I guess that i got a dump in the MOD. How can i solve that problem?? Any ideas??

Regards

Miguel

[Former Member]

Hi

You can use the float number (type F) instead of integer

Max

[Former Member]

Hello, it doesnt work with type F.

Any ideas??

[Former Member]

Hi

I've tried to run this and it seems to work fine:

DATA:   SUMA TYPE F,
        MULT TYPE F,
        NUM  TYPE STRING.

DATA: NUM_F  TYPE F.

DATA: DV TYPE  C.

NUM = '0200000005500002133200000154700076945190'.

MOVE NUM TO NUM_F.

MULT = 2.
SUMA = 0.

WHILE NUM_F <> 0.
  SUMA = SUMA + NUM_F MOD 10 * MULT.
  NUM_F = NUM_F DIV 10.
  MULT = MULT + 1.
  IF MULT > 7.
    MULT = 2.
  ENDIF.
ENDWHILE.

NUM_F = 11 - SUMA MOD 11.
IF NUM_F = 10.
  DV ='K'.
ELSEIF NUM_F = 11.
  DV = '0'.
ELSE.
  MOVE NUM_F TO DV.
ENDIF.

WRITE DV.

Max

[Former Member]

Hi,

  DATA: suma TYPE f,
        num  TYPE f.

does work...

Kr,

m.

[Former Member]

Hi

look

check the result with this data:

value = '011004455010010010116000005670', result is = 4

value = '021704490010010120516000005000', result is = 3

more easy

value = '15292975', result is = 7

I need the string up to 40 numbers, with 30 is working and with less also. But a big number get dumps

i changed the program..everything, it doest work, thanks anyway, trying to fix this issue.

Regards

[Former Member]

Hi

thanks, but it doesnt work

[Former Member]

Yes,

and with '0200000005500002133200000154700076945190', it gives 0...

and with '02000000055000021332000001547000769451908711234001' it gives 5...

it even works with a 80 len string

kr,

m.

Edited by: Manu D'Haeyer on Sep 27, 2011 7:26 PM

[Former Member]

Hi Manu.

When i changed the data as you said

  DATA: suma TYPE f,
        mult TYPE i,
        num  TYPE f.

the result of value = '011004455010010010116000005670' it gives me DV = 5.0000000000000000E+00

And it should be DV = 4

Am I doing sth wrong???

Regards

Edited by: Miguel Alvear on Sep 27, 2011 7:30 PM

[Former Member]

DATA: RESULT TYPE I.

DATA: DV(10) TYPE  C.

NUM = '11004455010010010116000005670'.

MOVE NUM TO NUM_F.

MULT = 2.
SUMA = 0.

WHILE NUM_F <> 0.
  SUMA = SUMA + NUM_F MOD 10 * MULT.
  NUM_F = NUM_F DIV 10.
  MULT = MULT + 1.
  IF MULT > 7.
    MULT = 2.
  ENDIF.
ENDWHILE.

NUM_F = 11 - SUMA MOD 11.
IF NUM_F = 10.
  DV ='K'.
ELSEIF NUM_F = 11.
  DV = '0'.
ELSE.
  MOVE NUM_F TO RESULT.
  MOVE RESULT TO DV.
ENDIF.

WRITE DV.

I've changed the program in order to convert the result in a integer number before moving it into a char.

Now if I run the program 011004455010010010116000005670, the result is 2

Max

[Former Member]

Well I guess you loose some precision with your DIV statement...

Try with:

num = num / 10.

and then, after looping:

num = 11 - suma MOD 11.
num = ceil( num ).

But I must say I don't really understand what you are trying to do here

kr,

m.

Edited by: Manu D'Haeyer on Sep 27, 2011 7:36 PM

[Former Member]

Hi

Yes Manu is right

It's not clear the logic of your algoritm, so which result you need to get, all we can say it's use a number type float in order to avoid the dump.

Max

[Former Member]

Hi guys.

The logic of the algoritm is ok, it is working with data type i. The result DV is also ok.

But the problem is that the value if bigger 40 lenght numbers and then

1.- get dump overflow

2.- if i changed to type F, the result is different than if i do it with type i...

that's my issue.

Regards

[Former Member]

With '011004455010010010116000005670' and a CEIL statement it gives 4. Is that correct?

Damn I want to understand now...:)

Kr,

m.

[Former Member]

Hi,

Yes correct

value=011004455010010010116000005670, DV = 4

value=011157638010010093416000005000, DV = 0

value=010037221010010102116000008050, DV = 1

value=021704490010010120516000005000, DV = 3

value=021706259010010111516000005000, DV = 4

value=041979446010010133316000005000, DV = 1

All of those are ok with the type i, if you change it to float F, it changes...it doesnt have to change.

but the number that the logic will use is longer than 30 as i can use with type i,

if i put a value long 40, using type i, death....and if i change it to type F, also death, the result is not OK

Regards

Trhanks for the help

[Former Member]

this issue i thought that it was simple to solve, but it gets more and more complicated...

heheeh

check the first code that i put, the original with type i, use it with the values that i put as samples.

then change the data with type F, use the values...the result is wrong.

Regards

[Former Member]

Yes indeed results are different... strange algorithmn you've built up!

Anyway, a last idea perhaps, could you not split the string to work on them separately...?

Well, which you luck!

Kr,

m.

[Former Member]

Try with type DECFLOAT34...

edit: oh no... not correct either... different however... That's crasy, a result different per type

Kr,

m.

Edited by: Manu D'Haeyer on Sep 27, 2011 8:09 PM

[Former Member]

Hi

The problem is type F doesn't save all integer part, but always the first 16 or 17 number, I means so the result will be different, but probably (you need to use only integer number) you can simple use the type N.

I've changed my code using a type N long 50 char:

PARAMETERS: P(50) TYPE C.

DATA:   SUMA(50)  TYPE N,
        MULT(50)  TYPE N,
        NUM_C(50) TYPE N,
        NUM      TYPE STRING.

DATA:  DV TYPE STRING.

NUM = P.

MOVE NUM   TO NUM_C.

MULT = 2.
SUMA = 0.

WHILE NUM_C <> 0.
  SUMA = SUMA + NUM_C MOD 10 * MULT.
  NUM_C = NUM_C DIV 10.
  MULT = MULT + 1.
  IF MULT > 7.
    MULT = 2.
  ENDIF.
ENDWHILE.

NUM_C = 11 - SUMA MOD 11.
IF NUM_C = 10.
  DV ='K'.
ELSEIF NUM_C = 11.
  DV = '0'.
ELSE.
  MOVE NUM_C TO DV.
ENDIF.

WRITE DV.

I've run it with all values you've have posted:

value=011004455010010010116000005670, DV = 4
value=011157638010010093416000005000, DV = 0
value=010037221010010102116000008050, DV = 1
value=021704490010010120516000005000, DV = 3
value=021706259010010111516000005000, DV = 4
value=041979446010010133316000005000, DV = 1

And I get the result expected.

Now the problem could be to delete the leading zero in the result stored in DV, but this is simple to be solved

Max

[Former Member]

Hi Max,

Was the first thing I tested... It works yes, but not for a 40 length string, which was the issue...

I've nearly gone crazy with this, don't think there is a solution without changing the logic

Kr,

m.

[Former Member]

Hello guys

thanks a lot for your help and effort.

I was able to change the logic of the program and now it works even with 1000 characters and even more.

I'm posting the code, so you can try it

FUNCTION zcalculo_dv.
*"----------------------------------------------------------------------
*"*"Interfase local
*"  IMPORTING
*"     REFERENCE(CADENA) TYPE  STRING
*"  EXPORTING
*"     REFERENCE(DV) TYPE  C
*"----------------------------------------------------------------------
* Toma una cadena de numeros, solo numeros (sin .,"·%!$) y calcula el DV

  DATA: suma    TYPE i,
        mult    TYPE i,
        largo   TYPE i,
        num     type i,
        dig     type i.

  largo = strlen( cadena ).

  mult = 2.
  suma = 0.

  WHILE largo <> 0.
    largo = largo - 1.
    dig = cadena+largo(1).
    suma = suma + dig * mult.
    mult = mult + 1.
    IF mult > 7.
      mult = 2.
    ENDIF.

  ENDWHILE.

  num = 11 - suma MOD 11.
  IF num = 10.
    dv ='K'.
  ELSEIF num = 11.
    dv = '0'.
  ELSE.
    dv = num.
  ENDIF.

ENDFUNCTION.

rather than use the whole value, now I'm using just each character separately and it works.

Regards

[Former Member]

Hello

thanks for the help and the effort.

I changed the logic of the code, have a look it, now it works fine.

Regards

Miguel

[Former Member]

Nice Thanks for sharing!...

Kr,

Manu.