Fortran Programming Language :: which is efficient


	which is efficient Hi This is the code DO J1=1,SS DO I1=1,TT DO ISUMT3=1,MA SUMGMT(I1,J1,ISUMT3)=0.0 END DO END DO END DO i can write the following code in FORTRAN 90 as SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 now my question is which one is the efficeint way to do it? I think i know the answer, the second way is more efficient. Am i right ? And if yes how do i prove it? How can i prove this with this simple code that the second one is efficient ? regards - Hide quoted text - aeroguy wrote: > This is the code > DO J1=1,SS > DO I1=1,TT > DO ISUMT3=1,MA > SUMGMT(I1,J1,ISUMT3)=0.0 > END DO > END DO > END DO > i can write the following code in FORTRAN 90 as > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > now my question is which one is the efficeint way to do it? > I think i know the answer, the second way is more efficient. > Am i right ? And if yes how do i prove it? How can i prove this with > this simple code that the second one is efficient ? Whether you are right or not depends on what compiler you are using, what optimization switches you are using, and what sort of computer you're running the compiled code on. It is also likely to depend on whether this is the entirety of the array, or only part of it. And on whether or not the compiler knows the answer to that question when it's compiling this particular code. (For example, if this is in a subroutine that gets an assumed-shape array, it may not know.) And it's also likely to be important that that's 0.0 and not some other number, too. On some systems, setting a large swath of memory to all zero bits is faster than setting it to some particular value. In any case, the truth of the matter is that, in order to get any really meaningful answer to this question, you need to test it yourself. Write a program that does one of these bits of code over and over, thousands of times. Time how long it takes. Do the same with the other code, and compare the results. Alternately, compile the two codes using a compiler switch which dumps out the assembly-level instructions it creates, and look at them and calculate which one takes more time. (You should probably do the latter anyway, just to make sure that the compiler isn't optimizing things entirely away.) However, I will bet that, for things this simple, any reasonably good optimizing compiler will realize that these two bits of code do exactly the same thing, and thus will produce exactly the same result for either one. And thus the right answer is: Use the second one, because it's easier to read. Meanwhile, I think there's a more important question: Does this even matter? If your program is written in such a way that it spends a significant amount of time setting large arrays to zero, then chances are reasonably good that have an algorithm that's wasting time zeroing arrays that don't need to be zeroed, and you should fix the algorithm instead. And, if your program doesn't spend a lot of time zeroing out the arrays, then it doesn't make a hill of beans worth of difference if one method is 2% faster than the other. - Brooks -- The "bmoses-nospam" address is valid; no unmunging needed. - Hide quoted text - aeroguy wrote: > Hi > This is the code > DO J1=1,SS > DO I1=1,TT > DO ISUMT3=1,MA > SUMGMT(I1,J1,ISUMT3)=0.0 > END DO > END DO > END DO > i can write the following code in FORTRAN 90 as > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > now my question is which one is the efficeint way to do it? > I think i know the answer, the second way is more efficient. > Am i right ? And if yes how do i prove it? How can i prove this with > this simple code that the second one is efficient ? Possibly so, since you would have to turn on a loop-nest optimization option to get good performance with the first version. Did you mean SUMGMT(1:TT,1:SS,1:MA) ? The answer certainly depends on circumstances which go beyond the Fortran standard or the information presented here. - Hide quoted text - aeroguy wrote: > Hi > This is the code > DO J1=1,SS > DO I1=1,TT > DO ISUMT3=1,MA > SUMGMT(I1,J1,ISUMT3)=0.0 > END DO > END DO > END DO > i can write the following code in FORTRAN 90 as > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > now my question is which one is the efficeint way to do it? > I think i know the answer, the second way is more efficient. Most likely. The compiler can use the most efficient order to loop through the array. > Am i right ? And if yes how do i prove it? How can i prove this with > this simple code that the second one is efficient ? You can try to check it, you cannot prove it. There is no garantee that the compiler can not be smart enough to optimize your loops. aeroguy <sukhbinder.si@gmail.com> wrote: > DO J1=1,SS > DO I1=1,TT > DO ISUMT3=1,MA > SUMGMT(I1,J1,ISUMT3)=0.0 > END DO > END DO > END DO > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > now my question is which one is the efficeint way to do it? It depends, but you should also try: DO ISUMT3=1,MA DO J1=1,SS DO I1=1,TT SUMGMT(I1,J1,ISUMT3)=0.0 END DO END DO END DO -- pa at panix dot com - Hide quoted text - aeroguy wrote: > Hi > This is the code > DO J1=1,SS > DO I1=1,TT > DO ISUMT3=1,MA > SUMGMT(I1,J1,ISUMT3)=0.0 > END DO > END DO > END DO > i can write the following code in FORTRAN 90 as > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > now my question is which one is the efficeint way to do it? Almost certainly the second as presented, but that's only true because you have order of your loop indices reversed. If you get the loop indices right (loop on the rightmost index first, then the middle, then the left), it might be either. > Am i right ? And if yes how do i prove it? How can i prove this with > this simple code that the second one is efficient ? Profiling can do this. Put each version in a subroutine, call it a lot of times, be careful to have it be a situation where the compiler doesn't get smart and optimize the whole thing out, and you should be able to tell from the timing for the two subroutines. You can also manually time it with e.g. CPU_TIME. On Apr 24, 11:38 pm, aeroguy <sukhbinder.si@gmail.com> wrote: - Hide quoted text - > Hi > This is the code > DO J1=1,SS > DO I1=1,TT > DO ISUMT3=1,MA > SUMGMT(I1,J1,ISUMT3)=0.0 > END DO > END DO > END DO > i can write the following code in FORTRAN 90 as > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > now my question is which one is the efficeint way to do it? > I think i know the answer, the second way is more efficient. > Am i right ? And if yes how do i prove it? How can i prove this with > this simple code that the second one is efficient ? > regards Just set SS, TT and MA to some reasonably large numbers so that the execution time would be longe enough to ignore side effects and measure what happens. I bet you won't see any difference. But you might get a speed up if you use some sort of autoparalelizing and/or vectorizing features of your compiler/processor. Cheers, Victor. On Apr 25, 1:34 pm, Michel Olagnon <molag@ifremer-a-oter.fr> wrote: - Hide quoted text - > aeroguy wrote: > > Hi > > This is the code > > DO J1=1,SS > > DO I1=1,TT > > DO ISUMT3=1,MA > > SUMGMT(I1,J1,ISUMT3)=0.0 > > END DO > > END DO > > END DO > > i can write the following code in FORTRAN 90 as > > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > > now my question is which one is the efficeint way to do it? > > I think i know the answer, the second way is more efficient. > Most likely. The compiler can use the most efficient order to loop > through the array. > > Am i right ? And if yes how do i prove it? How can i prove this with > > this simple code that the second one is efficient ? > You can try to check it, you cannot prove it. There is no garantee > that the compiler can not be smart enough to optimize your loops.- Hide quoted text - > Thnak you all for your answers and time. Well the code i am modifying is setting lot of arrays to zeros. so i will look to change the algorithm. Meanwhile i am not looking for any compiler optimisation. What i want is is there some ways in which by some user techniques and by using particular standards of FORTRAN one can get a optimised code and thus easy the work of a coomplier as such. I google and found lots of terms like loop interchange, loop tilting, loop blocking, and many others. I want to know is this techniques worth the effort. i will now list a subroutine of the program, can you please suggest a better, effecient way to remove/replace the GOTO jumble that it has: SUBROUTINE iskokaro1 (j,mde,inl) use wjans, only: lim1 use modul5, only: sndi,ssrec use modul29, only: v29 use modul85, only: noblm use lmodul6, only: un INTEGER, INTENT(IN) :: j INTEGER, INTENT(IN) :: mde INTEGER, INTENT(IN OUT) :: inl SAVE DIMENSION x(20(lim1+2)) INTEGER :: wjan,wjan0,wjan1,wjan2,blkno,filea,fileb INTEGER :: filec IF (inl == 0) GO TO 1 IF (inl > 1) GO TO 70 noblm=lim1 wjan=(noblm+2)20 ssndi=j IF (sndi == 0) sndi=1 wjan0=wjan/sndi-2 j1=1 j2=wjan0 wjan1=wjan0+1 wjan2=wjan0+2 jx=wjan2 nblk=0 blkno=1 kk=sndi initl=1 iwrite=0 filea=un(10) fileb=un(10) REWIND filea DO k=1,kk DO i=1,wjan2 x((i-1)kk+k)=0. END DO END DO DO k=1,kk v29(k)=0. END DO RETURN 1 kya=mde jj=j IF (kya == 0) GO TO 3 IF (jj /= 1) GO TO 3 IF (iwrite == 1) GO TO 3 IF (initl == 1) GO TO 3 iwrite=1 fileb=un(13) REWIND fileb 3 IF (jj >= j1) GO TO 25 nblk=MAX0(nblk,blkno) IF (kya == 0) GO TO 20 initl=0 DO k=1,kk x((jx-1)kk+k)=v29(k) END DO IF (blkno == 1) GO TO 18 ssrec=jx DO k=1,kk WRITE (fileb) (x((i-1)kk+k),i=1,ssrec) END DO IF (iwrite == 0) GO TO 10 REWIND filea ii=filea filea=fileb fileb=ii 10 REWIND filea DO k=1,kk READ (filea) (x((i-1)kk+k),i=1,wjan1) END DO j1=1 j2=x((wjan1-1)kk+1) blkno=1 18 jx=wjan2 RETURN 20 IF (blkno == 1) GO TO 18 GO TO 10 25 IF (jj > j2) GO TO 45 IF (kya == 0) GO TO 35 DO k=1,kk x((jx-1)kk+k)=v29(k) END DO IF (kya == 2) last=jx 35 jx=jj-j1+1 DO k=1,kk v29(k)=x((jx-1)kk+k) END DO RETURN 45 IF (kya == 0) GO TO 55 DO k=1,kk x((jx-1)kk+k)=v29(k) END DO IF (kya == 2) last=jx IF (initl == 1) x((wjan1-1)kk+1)=last-1 DO k=1,kk WRITE (fileb) (x((i-1)kk+k),i=1,wjan1) END DO IF (initl == 0) GO TO 55 jjj=jx-last+1 DO jj=1,jjj jx=last+jj-1 DO k=1,kk x((jj-1)kk+k)=x((jx-1)kk+k) END DO END DO jx=jjj+1 55 j1=j1+x((wjan1-1)kk+1) j2=j1+wjan0-1 blkno=blkno+1 IF (initl == 1) RETURN ii=wjan1 IF (blkno == nblk) ii=ssrec DO k=1,kk READ (filea) (x((i-1)kk+k),i=1,ii) END DO IF (blkno == nblk) x((wjan1-1)kk+1)=ssrec j2=j1+x((wjan1-1)kk+1)-1 GO TO 35 70 CONTINUE IF (j == 1) THEN filec=un(20) ELSE filec=un(33) END IF IF (inl == 2) THEN REWIND filec WRITE (filec) sndi,wjan0,wjan1,wjan2,nblk,ssrec DO n=1,nblk ii=wjan1 IF (n == nblk.AND.nblk > 1) ii=ssrec DO k=1,kk IF (n > 1) READ (filea) (x((i-1)kk+k),i=1,ii) WRITE (filec) (x((i-1)kk+k),i=1,ii) END DO END DO ELSE READ (filec) sndi,wjan0,wjan1,wjan2,nblk,ssrec kk=sndi initl=0 iwrite=0 filea=un(10) REWIND filea ssrec=0 DO n=1,nblk ii=wjan1 IF (n == nblk.AND.nblk > 1) ii=ssrec ssrec=ssrec+ii-1 DO k=1,kk READ (filec) (x((i-1)kk+k),i=1,ii) IF (nblk > 1) WRITE (filea) (x((i-1)kk+k),i=1,ii) END DO END DO END IF REWIND filec IF (nblk > 1) THEN GO TO 10 ELSE GO TO 18 END IF RETURN END SUBROUTINE iskokaro1 Because of the time gap, we cant't have live commenting on this post but if you have question , please mention it will reply on next day. thanks for your time and consideration On Apr 25, 1:34 pm, Michel Olagnon <molag@ifremer-a-oter.fr> wrote: - Hide quoted text - > aeroguy wrote: > > Hi > > This is the code > > DO J1=1,SS > > DO I1=1,TT > > DO ISUMT3=1,MA > > SUMGMT(I1,J1,ISUMT3)=0.0 > > END DO > > END DO > > END DO > > i can write the following code in FORTRAN 90 as > > SUMGMT(1:TT,1:SS,1:ISUMT3) = 0.0 > > now my question is which one is the efficeint way to do it? > > I think i know the answer, the second way is more efficient. > Most likely. The compiler can use the most efficient order to loop > through the array. > > Am i right ? And if yes how do i prove it? How can i prove this with > > this simple code that the second one is efficient ? > You can try to check it, you cannot prove it. There is no garantee > that the compiler can not be smart enough to optimize your loops.- Hide quoted text - > Thnak you all for your answers and time. Well the code i am modifying is setting lot of arrays to zeros. so i will look to change the algorithm. Meanwhile i am not looking for any compiler optimisation. What i want is is there some ways in which by some user techniques and by using particular standards of FORTRAN one can get a optimised code and thus easy the work of a coomplier as such. I google and found lots of terms like loop interchange, loop tilting, loop blocking, and many others. I want to know is this techniques worth the effort. i will now list a subroutine of the program, can you please suggest a better, effecient way to remove/replace the GOTO jumble that it has: SUBROUTINE iskokaro1 (j,mde,inl) use wjans, only: lim1 use modul5, only: sndi,ssrec use modul29, only: v29 use modul85, only: noblm use lmodul6, only: un INTEGER, INTENT(IN) :: j INTEGER, INTENT(IN) :: mde INTEGER, INTENT(IN OUT) :: inl SAVE DIMENSION x(20(lim1+2)) INTEGER :: wjan,wjan0,wjan1,wjan2,blkno,filea,fileb INTEGER :: filec IF (inl == 0) GO TO 1 IF (inl > 1) GO TO 70 noblm=lim1 wjan=(noblm+2)20 ssndi=j IF (sndi == 0) sndi=1 wjan0=wjan/sndi-2 j1=1 j2=wjan0 wjan1=wjan0+1 wjan2=wjan0+2 jx=wjan2 nblk=0 blkno=1 kk=sndi initl=1 iwrite=0 filea=un(10) fileb=un(10) REWIND filea DO k=1,kk DO i=1,wjan2 x((i-1)kk+k)=0. END DO END DO DO k=1,kk v29(k)=0. END DO RETURN 1 kya=mde jj=j IF (kya == 0) GO TO 3 IF (jj /= 1) GO TO 3 IF (iwrite == 1) GO TO 3 IF (initl == 1) GO TO 3 iwrite=1 fileb=un(13) REWIND fileb 3 IF (jj >= j1) GO TO 25 nblk=MAX0(nblk,blkno) IF (kya == 0) GO TO 20 initl=0 DO k=1,kk x((jx-1)kk+k)=v29(k) END DO IF (blkno == 1) GO TO 18 ssrec=jx DO k=1,kk WRITE (fileb) (x((i-1)kk+k),i=1,ssrec) END DO IF (iwrite == 0) GO TO 10 REWIND filea ii=filea filea=fileb fileb=ii 10 REWIND filea DO k=1,kk READ (filea) (x((i-1)kk+k),i=1,wjan1) END DO j1=1 j2=x((wjan1-1)kk+1) blkno=1 18 jx=wjan2 RETURN 20 IF (blkno == 1) GO TO 18 GO TO 10 25 IF (jj > j2) GO TO 45 IF (kya == 0) GO TO 35 DO k=1,kk x((jx-1)kk+k)=v29(k) END DO IF (kya == 2) last=jx 35 jx=jj-j1+1 DO k=1,kk v29(k)=x((jx-1)kk+k) END DO RETURN 45 IF (kya == 0) GO TO 55 DO k=1,kk x((jx-1)kk+k)=v29(k) END DO IF (kya == 2) last=jx IF (initl == 1) x((wjan1-1)kk+1)=last-1 DO k=1,kk WRITE (fileb) (x((i-1)kk+k),i=1,wjan1) END DO IF (initl == 0) GO TO 55 jjj=jx-last+1 DO jj=1,jjj jx=last+jj-1 DO k=1,kk x((jj-1)kk+k)=x((jx-1)kk+k) END DO END DO jx=jjj+1 55 j1=j1+x((wjan1-1)kk+1) j2=j1+wjan0-1 blkno=blkno+1 IF (initl == 1) RETURN ii=wjan1 IF (blkno == nblk) ii=ssrec DO k=1,kk READ (filea) (x((i-1)kk+k),i=1,ii) END DO IF (blkno == nblk) x((wjan1-1)kk+1)=ssrec j2=j1+x((wjan1-1)kk+1)-1 GO TO 35 70 CONTINUE IF (j == 1) THEN filec=un(20) ELSE filec=un(33) END IF IF (inl == 2) THEN REWIND filec WRITE (filec) sndi,wjan0,wjan1,wjan2,nblk,ssrec DO n=1,nblk ii=wjan1 IF (n == nblk.AND.nblk > 1) ii=ssrec DO k=1,kk IF (n > 1) READ (filea) (x((i-1)kk+k),i=1,ii) WRITE (filec) (x((i-1)kk+k),i=1,ii) END DO END DO ELSE READ (filec) sndi,wjan0,wjan1,wjan2,nblk,ssrec kk=sndi initl=0 iwrite=0 filea=un(10) REWIND filea ssrec=0 DO n=1,nblk ii=wjan1 IF (n == nblk.AND.nblk > 1) ii=ssrec ssrec=ssrec+ii-1 DO k=1,kk READ (filec) (x((i-1)kk+k),i=1,ii) IF (nblk > 1) WRITE (filea) (x((i-1)kk+k),i=1,ii) END DO END DO END IF REWIND filec IF (nblk > 1) THEN GO TO 10 ELSE GO TO 18 END IF RETURN END SUBROUTINE iskokaro1 Because of the time gap, we cant't have live commenting on this post but if you have question , please mention it will reply on next day. thanks for your time and consideration "aeroguy" <sukhbinder.si@gmail.com> wrote in message news:1177564368.272312.142740@n15g2000prd.googlegroups.com... > On Apr 25, 1:34 pm, Michel Olagnon <molag@ifremer-a-oter.fr> wrote: [liberal editting] - Hide quoted text - > Thnak you all for your answers and time. Well the code i am modifying > is setting lot of arrays to zeros. so i will look to change the > algorithm. Meanwhile i am not looking for any compiler optimisation. > What i want is is there some ways in which by some user techniques and > by using particular standards of FORTRAN one can get a optimised code > and thus easy the work of a coomplier as such. > I google and found lots of terms like loop interchange, loop tilting, > loop blocking, and many others. I want to know is this techniques > worth the effort. > i will now list a subroutine of the program, can you please suggest a > better, effecient way to remove/replace the GOTO jumble that it has: > SUBROUTINE iskokaro1 (j,mde,inl) > use wjans, only: lim1 > use modul5, only: sndi,ssrec > use modul29, only: v29 > use modul85, only: noblm > use lmodul6, only: un > Because of the time gap, we cant't have live commenting on this post > but if you have question , please mention it will reply on next day. > thanks for your time and consideration On my server, it shows this message twice. If this was your source, you have a demonstrably subtle mind. Did you post the message twice or did Google do that for you? -- WW - Hide quoted text - aeroguy wrote: > i will now list a subroutine of the program, can you please suggest a > better, effecient way to remove/replace the GOTO jumble that it has: > SUBROUTINE iskokaro1 (j,mde,inl) > use wjans, only: lim1 > use modul5, only: sndi,ssrec > use modul29, only: v29 > use modul85, only: noblm > use lmodul6, only: un > INTEGER, INTENT(IN) :: j > INTEGER, INTENT(IN) :: mde > INTEGER, INTENT(IN OUT) :: inl > SAVE > DIMENSION x(20(lim1+2)) > INTEGER :: wjan,wjan0,wjan1,wjan2,blkno,filea,fileb > INTEGER :: filec > IF (inl == 0) GO TO 1 > IF (inl > 1) GO TO 70 > noblm=lim1 > wjan=(noblm+2)20 > ssndi=j > IF (sndi == 0) sndi=1 > wjan0=wjan/sndi-2 > j1=1 > j2=wjan0 > wjan1=wjan0+1 > wjan2=wjan0+2 > jx=wjan2 > nblk=0 > blkno=1 > kk=sndi > initl=1 > iwrite=0 > filea=un(10) > fileb=un(10) > REWIND filea > DO k=1,kk > DO i=1,wjan2 > x((i-1)kk+k)=0. > END DO > END DO > DO k=1,kk > v29(k)=0. > END DO > RETURN > 1 kya=mde > jj=j > IF (kya == 0) GO TO 3 > IF (jj /= 1) GO TO 3 > IF (iwrite == 1) GO TO 3 > IF (initl == 1) GO TO 3 > iwrite=1 > fileb=un(13) > REWIND fileb > 3 IF (jj >= j1) GO TO 25 > nblk=MAX0(nblk,blkno) > IF (kya == 0) GO TO 20 > initl=0 > DO k=1,kk > x((jx-1)kk+k)=v29(k) > END DO > IF (blkno == 1) GO TO 18 > ssrec=jx > DO k=1,kk > WRITE (fileb) (x((i-1)kk+k),i=1,ssrec) > END DO > IF (iwrite == 0) GO TO 10 > REWIND filea > ii=filea > filea=fileb > fileb=ii > 10 REWIND filea > DO k=1,kk > READ (filea) (x((i-1)kk+k),i=1,wjan1) > END DO > j1=1 > j2=x((wjan1-1)kk+1) > blkno=1 > 18 jx=wjan2 > RETURN > 20 IF (blkno == 1) GO TO 18 > GO TO 10 > 25 IF (jj > j2) GO TO 45 > IF (kya == 0) GO TO 35 > DO k=1,kk > x((jx-1)kk+k)=v29(k) > END DO > IF (kya == 2) last=jx > 35 jx=jj-j1+1 > DO k=1,kk > v29(k)=x((jx-1)kk+k) > END DO > RETURN > 45 IF (kya == 0) GO TO 55 > DO k=1,kk > x((jx-1)kk+k)=v29(k) > END DO > IF (kya == 2) last=jx > IF (initl == 1) x((wjan1-1)kk+1)=last-1 > DO k=1,kk > WRITE (fileb) (x((i-1)kk+k),i=1,wjan1) > END DO > IF (initl == 0) GO TO 55 > jjj=jx-last+1 > DO jj=1,jjj > jx=last+jj-1 > DO k=1,kk > x((jj-1)kk+k)=x((jx-1)kk+k) > END DO > END DO > jx=jjj+1 > 55 j1=j1+x((wjan1-1)kk+1) > j2=j1+wjan0-1 > blkno=blkno+1 > IF (initl == 1) RETURN > ii=wjan1 > IF (blkno == nblk) ii=ssrec > DO k=1,kk > READ (filea) (x((i-1)kk+k),i=1,ii) > END DO > IF (blkno == nblk) x((wjan1-1)kk+1)=ssrec > j2=j1+x((wjan1-1)kk+1)-1 > GO TO 35 > 70 CONTINUE > IF (j == 1) THEN > filec=un(20) > ELSE > filec=un(33) > END IF > IF (inl == 2) THEN > REWIND filec > WRITE (filec) sndi,wjan0,wjan1,wjan2,nblk,ssrec > DO n=1,nblk > ii=wjan1 > IF (n == nblk.AND.nblk > 1) ii=ssrec > DO k=1,kk > IF (n > 1) READ (filea) (x((i-1)kk+k),i=1,ii) > WRITE (filec) (x((i-1)kk+k),i=1,ii) > END DO > END DO > ELSE > READ (filec) sndi,wjan0,wjan1,wjan2,nblk,ssrec > kk=sndi > initl=0 > iwrite=0 > filea=un(10) > REWIND filea > ssrec=0 > DO n=1,nblk > ii=wjan1 > IF (n == nblk.AND.nblk > 1) ii=ssrec > ssrec=ssrec+ii-1 > DO k=1,kk > READ (filec) (x((i-1)kk+k),i=1,ii) > IF (nblk > 1) WRITE (filea) (x((i-1)kk+k),i=1,ii) > END DO > END DO > END IF > REWIND filec > IF (nblk > 1) THEN > GO TO 10 > ELSE > GO TO 18 > END IF > RETURN > END SUBROUTINE iskokaro1 Go back to original compilable code and straighten that out before inserting new f90 syntax.