JP- What you're really saying in your x1=x2=x3=x4=0 is: x1 = (x2=x3=x4=0) where "x2=x3=x4=0" is a logical expression that will return false (0). The expression is basically asking "are x2, x3, x4 and 0 all the same?" At this point, since x2-x4 have never been referenced, they are not 0 and so the expression is false. You're right in your original code - in this case: x1=0; x2=0; x3=0; x4=0;

Hope this helps- Pete

home: pete.lund@home.com

On Thu, 09 Nov 2000 21:35:31 GMT, "JP" <R_E_M_O_V_E_lecruguel@epid.jgh.mcgill.ca> wrote:

> >Hi all > >I tried >pbhbinc= pbhbinc= pbhbinc2= 0; >vs. >pbhbinc= 0; pbhbinc0=0; pbhbinc2= 0; >thinking it would do the same. > >As I run an ARRAY statement later. I thought that I would get the same in >the end. >As it was just an initialisation of the 3 variables. > >Instead I got: pbhbinc ok but pbhbinc0 & pbhbinc2 ate set to missing for >all obs. > >How does the x1=x2 = x3=x4 = 0 work? what's the logic? >Should it be used in some case? > >JP > >here is the code: > >DATA c2b; > SET c2a; > ARRAY sjr{*} jour1-jour731; > > ARRAY dmdt{*} ddde1-ddde8; > ARRAY dmtp{*} Typeheb1-Typeheb8; > ARRAY dmf{*} dferm1-dferm8; > > ARRAY hbdta{*} dadm1-dadm18; > ARRAY hbdts{*} dsor1-dsor18; > ARRAY hbtp{*} typea1-typea18; > ARRAY hbrd{*} raisond1-raisond18; > > ARRAY hpdta{*} datadm1-datadm35; > ARRAY hpdts{*} datsor1-datsor35; > * met HEB non fini a '01APR2000'd; > * met HEB = HOP parfait à HEB = .; > * détecte chevauchement HEB/HOP, HEB inclus dans HOP, HOP includ dans HEB >mme jour adm ou me jour sortie; >pbhbinc= 0; pbhbinc0=0; pbhbinc2= 0; > > DO j = 1 TO DIM(hbdta); > IF hbdta{j} AND hbdts{j} = . THEN hbdts{j} = '01APR2000'd; > DO k = 1 TO DIM(hpdta);*date sortie heb vide mise a avril 2000; > IF hbdta{j} AND hpdta{k} AND (hbdta{j} = hpdta{k} AND hbdts{j} = >hpdts{k}) THEN DO;hbdta{j} =.; hbdts{j} =.;END; *heb = hop heb mis a vide; > IF hbdta{j} AND hpdta{k} AND ((hbdta{j} >= hpdta{k} AND hbdts{j} >= >hpdts{k} AND hbdta{j} < hpdts{k}) > OR ( hbdta{j} =< hpdta{k} AND hbdts{j} =< hpdts{k} AND hbdts{j} >> hpdta{k} )) THEN pbhbinc0 = pbhbinc0+1; *chevauchement; > IF hbdta{j} AND hpdta{k} AND hbdta{j} >= hpdta{k} AND hbdts{j}=< >hpdts{k} THEN pbhbinc = pbhbinc+1; *heb inclus dans hosp; > IF hbdta{j} AND hpdta{k} AND (hbdta{j} = hpdta{k} OR hbdts{j} = >hpdts{k}) THEN pbhbinc2 = pbhbinc2+1; *entrée hosp = entrée heb ou sortie >hosp = sortie heb; > END; > END; > >RUN; >