Even number routine

[Former Member]

I have to test if a value is even. The value can be from 0 - 560. What is the simpliest (and most efficient) way of writing this routine

thanks

[Former Member]

hi,

The best and easiest method..

Do 560 times.

lval = sy-index mod 2.

if lval EQ '0'.

• EVEN

ELSE.

*ODD

ENDIF.

ENDIF.

[Mohamed_Mukhtar]

hi,

DATA : num TYPE i,
       rem TYPE i.

DO 560 TIMES.
  rem = sy-index MOD 2.
  IF rem = 0.
    WRITE : / sy-index,  '-even'.
  ELSE.
    WRITE : / sy-index, '-odd'.
  ENDIF.
ENDDO.

Thanks & Regards

[Former Member]

The mod function is what you need.

DATA: mod_value TYPE i.

mod_value = 21 MOD 2 .

IF mod_value EQ '0'.
  WRITE / 'even'.
ELSE.
  WRITE / 'odd'.
ENDIF.

[Former Member]

Hi Dominic ,

the efficient way is :

parameter : number .

DATA: mod TYPE i.

mod = number MOD 2 .

IF mod EQ '0'.

WRITE / 'Number is even'.

ELSE.

WRITE / ''Number is odd'.

ENDIF.

It will work most efficiently .

Regards ,

Nilesh Jain .

[Former Member]

Hi,

the most efficient way to do this is

data : n type i.

do 280 times.

n = 2 * sy-index.

write 😕 'even', n.

enddo.

this is the shortest code and will do fast.

hope this will help you.

regards,

anand

[Former Member]

REPORT Z_N_EVEN_NO.

DATA: V_NO TYPE I,

       RESULT TYPE I.

PARAMETERS: P_NO TYPE I.

IF RESULT = P_NO MOD 2.

   RESULT = 0.

   WRITE:/ ' THE NO  IS EVEN NO'.

   ELSE.

     WRITE:/ ' THE NO IS AN ODD NO'.

ENDIF.