is this code correct??

roshanvarghese · New User Joined: 15 Jul 2005 Posts: 1

hi

cud u pls explain how the call works in this case and also tell me the o/p

test: procedure options(main);

declare a fixed bin(31);

a = 100;
call subroutine(a,100);
put skip list (a,100);

end test;

subroutine: procedure(x,y);
declare (x,y) fixed bin(31);

x = 101;
y = 101;

end subroutine;

thanks
roshan

bakarthikeyan · New User Joined: 21 Sep 2005 Posts: 6

No. you have to use DCL <EXTERNAL PROC NAME> EXTERAL; in the calling program. So that, u can combine both.

Sridevi_C · Posted: Thu Sep 22, 2005 6:50 pm

Yep, external proc name must be declared.
The output will be 101 and 100.
Do correct me for any mistakes.
Thanks!
Sridevi.

jawadshaik · New User Joined: 27 Sep 2005 Posts: 16

proc1: proc options(main);

dcl x bin fixed(15) init(10);

dcl sub_proc entry(fixed bin(15) , fixed bin(15));

put skip list(x,a);

end proc1;

sub_proc: proc(x,y);

dcl (x,y) fixed bin(15);

x = x + 1;

y = y + 1;

end sub_proc;

output:

101

100

hope this will be clear

regards,

jawad shaik

jawadshaik · New User Joined: 27 Sep 2005 Posts: 16

SORRY FRIEND

CALL SUB_PROC(X,100); IS MISSED AFTER

SO

dcl sub_proc entry(fixed bin(15) , fixed bin(15));

call sub_proc(x,100);

put skip list(x,100);

AND OUTPUT IS

11

100

sorry for mistake plz dont mind

Sridevi_C · Posted: Wed Oct 19, 2005 8:28 pm

Hi jawadshaik,
Roshan wants to know the output for HIS code. YOUR code is different,fine. Even after considering your 2nd post, I couldn't predict what you are trying to convey. In your 1st code,X is initialized to 10 but the output is 101 and 100.What is variable "a"? Your 2nd code's output is 11 and 100. Please be clear on what you would like to convey.
Regards,
Sridevi.

jawadshaik · New User Joined: 27 Sep 2005 Posts: 16

Hi,

Sorry For Confusing

Here is what I meant

PROC1: PROC OPTIONS(MAIN);

DCL X FIXED BIN(15) STATIC INIT(100);

DCL SUB_PROC ENTRY(FIXED BIN(15),FIXED BIN(15));

CALL SUB_PROC(X,100);

PUT SKIP LIST(X,100);

END PROC1;

SUB_PROC: PROC(Y,Z);

DCL (Y,Z) FIXED BIN(15);

Y = Y + 1;

Z = Z + 1;

END SUB_PROC;

OUTPUT:

101 100

Hope I am clear know

regards,

Jawad shaik