du fortran au turbo pascal 7

Question

Bonsoir, je suis débutant en programation, j'ai deux codes en fortran que je veux faire en turbo pascal 7 les deux codes sont pour le calcul matriciel, le premeir c'est la méthode des approximations successives et le second est la méthode de hardy-cross! 
Je vous demande de l'aide 
mon email : messaoudi_choukri@yahoo.fr
Merci
le premier code:
        logical IPRINT, PGIVEN, INCID, CONV
        real L, NUMER
        dimension P(10), PGIVEN(10), A(10,10), C(10,10), D(10,10), 
        1    L(10,10), INCID(10,10), Q(10,10)
        *****************************************
        *****    Lecture des Données    ***** 
        *****************************************
1       read (5,100) N, IMAX, RHO, EPS, F, IPRINT, (PGIVEN(i), i = 1,N)
         read (5,101) (P(i), i = 1,N)
         do 2   i = 1,N
         read (5,102) (INCID(i,,j), j = 1,i)
         read (5,101) (D(i,j), j = 1,i)
2       read (5,101) (L(i,j), j = 1,i)
        *****************************************
        *****   transformation des matrices symetriques D, L et INCID et 
                 calcul des elements de la matice C   *****
        *****************************************
        FACTOR = 8.*12.**5*RHO*F/(3.1415926**2*32.2*(60.*7.48)**2*144.)
        do 3   i = 1,N
        do 3   j = 1,i
        C(i,j) = 0.
        if (i.EQ.j) go to 3
        if (INCID(i,j))  C(i,j) = FACTOR*L(i,j)/D(i,j)**5
        D(j,i) = D(i,j)
        L(j,i) = L(i,j)
        INCID(j,i) = INCID(i,j)
        C(j,i) = C(i,j)
3       continue
         *****************************************
         *****  ecriture des données initiales du reseau   *****
         *****************************************
        write (6,200) N,ITMAX,RHO,EPS,F,IPRINT,(i, P(i),PGIVEN(i),i=1,N)    
        write (6,201) 
        do 4    i = 1,N
4      write (6,202) i, i, N, (INCID(i,j), j = 1,N)
        write (6,201)
        do 5   i = 1,N
5      write (6,203) i, i, N, (D(i,j), j = 1,N)
        write (6,201)
        do 6   i = 1,N
6      write (6,204) i, i, N, (L(i,j), j = 1,N)
        *****************************************
        *****   calcul des approximations successives des pressions aux
                 noeuds   *****
        ****************************************
        if (IPRINT) write (6,205) (i, i = 1,N)
        do 9   ITER = 1, IMAX
        CONV = .true.
        do 8   j = 1, N
        if (PGIVEN) go to 8
        NUMER = 0.
        DENOM = 0.
        do 7   i = 1, N
        if (.not. INCID(i,j)) go to 7
           A(i,j) = 1.0/SQRT(C(i,j)*ABS(P(i)-P(j)))
           NUMER = NUMER + A(i,j)*P(i)
           DEMON = DEMON + A(i,j)
7      continue
        SAVEP = P(j)
        P(j) = NUMER/DENOM
        if (ABS(SAVEP-P(j)).GE.EPS) CONV = .false.
8      continue
        if (IPRINT) write (6,206) ITER, (P(i), i = 1,N)
        if (CONV) go to 10
9      continue
        write (6,207) ITMAX
        *****************************************
        *****  calcul des debits dans chaque troncon   *****
        *****************************************
10     do 11   i = 1,N
        do 11   j = 1,i
        Q(i,j) = 0.
        Q(j,i) = 0.
        if (i.EQ.j.OR..not.INCID(i,j)) go to 11
        Q(i,j) = (P(i)-P(j))/SQRT(C(i,j)*ABS(P(i)-P(j)))
        Q(j,i) = -Q(i,j)
11     continue
         ****************************************
         *****  ecriture des pressions finales et des debits  *****
         ****************************************
         write (6,208) ITER, N
         do 12  i = 1,N
12      write (6,209) i, P(i), (Q(i,j), j = 1,N)
         go to 1
         ***************************************
         ***** formats des etats d'entrées/sorties *****
         ***************************************
100      format(3x,i2,17x,i3,15x,F5.1,15x,E5.0/4x,F6.3,14x,L1 /
          1     (30x,20(L1,1x)))
101      format(30x,5F8.3)
102      format(30x,20(L1,1x))
200      format(23H1FLOW IN A PIPE NETWORK/ 10H0N = ,13/ 10H ITMAX
          1=, i3/ 10H RHO      =, F7.3/ 10H EPS     =, E10.2/ 10H F    =,
          2 F7.3/10H IPRINT =, 2x, L1/ 3H0 i, 6x, 4HP(i), 4x,9HPGIVEN(i)
          3 (1H, i2, F10.3, 6x, L1))                                                        
201      format(1H0/1H0)
202      format(7H0INCID(,i2,13H, 1)...INCID(, i2, 1H,,i2, 3H) = ,
          1 40(L1, 1x)/ (1H, 29x, 40(L1, 1x)))
203      format(3H0D(,i2, 9H, 1)...D(,i2,1H,,i2, 1H), 9x, 1H= , 8F10.3 /
          1 (1H, 29x, 8F10.3))
204      format(3H0L(,i2, 9H, 1)...L(,i2,1H,,i2, 1H), 9x, 1H= , 8F10.3 /
          1 (1H, 29x, 8F10.3))   
205      format(1H0/ 5H0ITER,7x,16HPRESSURE AT NODE/ (1H,11x,8(i1,9x)))  
206      format(1H, i3, 3x, 8F10.4/ (1H , 6x, 8F10.4))
207     format(35H0SOLUTIONS FAILED TO CONVERGE AFTER,i3,11H ITERATIONS) 
208     format(1H0/26H0PRESSURES AND FLOWS AFTER, i3,15H ITERATIONS ARE/ 
          1 3H0 i,5x,4HP(i),7x,16HQ(i, 1)...Q(i,,i2,1H) / 1H , 7x, 3HPSI,
          2 14x, 7HGAL/MIN//)
209     format(1H , i2, F10.4, 5x, 8F10.3/ (1H , 17x, 8F10.3))
       end

le second code :

C************************************************************************
C                        HARDY CROSS ANALYSIS II
C     THIS PROGRAM ANALYZES A PIPE NETWORK BY THE HARDY CROSS METHOD
C************************************************************************
C     NUML = NUMBER OF LOOPS
C     NUMPI = TOTAL NUMBER OF PIPES
C     EPS = EPSILON THE ACCEPTABLE ERROR
C     L IF THE NUMBER OF PIPES IN A LOOP. THE VALUE IS OBTAINED FROM J(L)
C     ID(L,K) IS AN INDEX WHICH ENABLES NUMBERING THE PIPES IN EACH LOOP
C     FROM ONE TO THE NUMBER OF PIPES IN THE LOOP WHILE STILL RETAINING
C     THE ORIGINAL PIPE NUMBERING SCHEME
C     ID(L,K)IS AN INDEX WHICH INCLUDES THE PIPE NUMBER AND A SIGN H ,
C     THE SIGN IS POSITIVE FOR ALL PIPES NOT COMMON TO MORE THAN ONE
C     LOOP. FOR PIPES FALLING IN TWO LOOPS, THE SIGN IS POSITIVE FOR THE
C     FIRST LOOP IN WHICH THE PIPE APPEARS AND NEGATIVE FOR THE SECOND
C     LOOP. FOR EXAMPLE, IF PIPE 3 IS COMMON TO LOOPS 1 AND 2, ID(1,3)C     +3 AND ID(2,1) -3 WHERE 3 IS THE THIRD PIPE IN LOOP 1 AND THE
C     FIRST PIPE IN LOOP 2
C     D IS THE PIPE DIAMETER IN FEET
C     Q(I) IS THE FLOW IN PIPE I (CFS)
C     THE OVERALL PIPE NETWORK IS BROKEN UP INTO L LOOPS
C     PL IS THE PIPE LENGTH IN THOUSANDS OF FEET. FOR A PIPE LENGTH OF
C     6,000 FEET, PL WOULD BE 6
C     S IS THE SLOPE IN FEET PER THOUSAND FEET. IT IS COMPUTED USING THE
C     HAZEB WILLAMS PIPE FORMULA. S = (Q/(0.0103*C*D**2.63))**1.85
C     WHERE C IS THE HAZEN WILLIAMS C
C     FOR VALUES OF C WHICH DIFFER FOR INDIVIDUAL PIPES A SIMPLE
C     PROGRAMMING CHANGE TO READ THE C VALUES IN AS DATA WOULD BE
C     SUFFICIENT
C     H IS THE HEAD LOSS IN EACH PIPE IN FEET H = S*PL
C     HOVQ IS THE RATIO H/Q
C     SHOVQ IS THE SUM OF THE RATIO H/Q
C     SUMH IS THE SUM OF THE HEAD LOSSES H
C     CORR IS THE CORRECTION TO BE APPLIED TO THE LOOPS
C     CORR = -SUMH/(1.85*SHOVQ)
C     SCOR IS THE SUM OF THE ABSOLUTE VALUES OF THE CORR(L)'S FOR EACH
C     LOOP. THIS VALUE IS COMPARED TO AN EPSILON. IF SCOR IS GREATER THAN
C     EPS, ANOTHER ITERATION IS MADE


C************************************************************************
      DIMENSION Q(400),D(400),PL(400),S(400),H(400),HOVQ(400),SHOVQ(60)
     .,SUMH(60),CORR(60),SQ(400),J(60),ID(60,120)
C************************************************************************
      CHARACTER*20 INP, OUT                             
      PRINT*,'ENTER INPUT FILE NAME :'
      READ*, INP
      PRINT*,'ENTER OUTPUT FILE NAME :'
      READ*, OUT
      OPEN(UNIT=5,FILE=INP)
      OPEN(UNIT=6,FILE=OUT)
C************************************************************************
      EPS=0.0001
      READ(5,37) NUML, NUMPI
   37 FORMAT(2I5)
      WRITE(6,39) NUML, NUMPI
   39 FORMAT(1X,2I5)
      READ(5,38) C
   38 FORMAT(F6.1)
      WRITE(6,40) C
   40 FORMAT(1X,F6.1)
      READ(5,20)(J(L),L=1,NUML)
   20 FORMAT(4I5)
      WRITE(6,36)(J(L),L=1,NUML)
   36 FORMAT(1X,4I5)
      READ(5,21)(D(I),Q(I),PL(I),I=1,NUMPI)
   21 FORMAT(9F6.3)
      WRITE(6,29)(D(I),Q(I),PL(I),I=1,NUMPI)
   29 FORMAT(1X,9F6.3)
      DO 23 L=1,NUML
      M=J(L)
      DO 23 K=1,M
      READ(5,24)ID(L,K)
   23 WRITE(6,35)ID(L,K)
   24 FORMAT(I5)
   35 FORMAT(1X,I5)
      R=C*0.0103
   17 SCOR=0.
      DO 1 I=1,NUMPI
      IF(Q(I).LT.0.) SQ(I)=-Q(I)
      IF(Q(I).GE.0.) SQ(I)=Q(I)
      S(I)=(SQ(I)/(R*D(I)**2.63))**1.85
      H(I)=S(I)*PL(I)
    1 HOVQ(I)=H(I)/SQ(I)
      DO 16 L=1,NUML
      SUMH(L)=0.
      SHOVQ(L)=0.
      M=J(L)
      DO 7 K=1,M
      IF(ID(L,K))3,3,4
    3 I=-ID(L,K)
      SHOVQ(L)=SHOVQ(L)+HOVQ(I)
      IF(Q(I))5,5,6
    4 I=ID(L,K)
      SHOVQ(L)=SHOVQ(L)+HOVQ(I)
      IF(Q(I))6,5,5
    6 SUMH(L)=SUMH(L)-H(I)
      GO TO 7
    5 SUMH(L)=SUMH(L)+H(I)
    7 CONTINUE
      CORR(L)=-SUMH(L)/(1.85*SHOVQ(L))
      M=J(L)
      DO 8 K=1,M
      IF(ID(L,K))10,10,11
   10 I=-ID(L,K)
      Q(I)=Q(I)-CORR(L)
      GO TO 8
   11 I=ID(L,K)
      Q(I)=Q(I)+CORR(L)
    8 CONTINUE
   16 SCOR=SCOR+ABS(CORR(L))
      IF(SCOR-EPS.GT.0.)GO TO 17
      WRITE(6,25)
   25 FORMAT(1H1,20X,15HTABLE OF VALUES,///)
      WRITE(6,26)
   26 FORMAT(7X,1H1,11X,4HQ(I),13X,4HS(I),8X,7HH(I)ABS,8X,5HPL(I),///)
      DO 27 I=1,NUMPI
   27 WRITE(6,28)I,Q(I),S(I),H(I),PL(I)
   28 FORMAT(6X,I2,3X,E14.4,3X,E14.4,3X,E14.4,3X,F6.3)
      STOP
      END

f0xi · Answer

Cher membre,

nous somme au regret de vous annoncer que nous n'assurons plus aucune aide pour les langages et compilateur sortis avant 2005.

nous ne pouvons donner suite à votre question.

nous laissons votre sujet ouvert tout de même au cas ou un fou furieux voudrais tout de même tenter l'exercice de cette translation Fortran->TP7.

Nous vous conseillons également de foutre un bon gros coups de latte dans votre 80486 (méthode certifié et approuvé), de le lourder au fin fond du désert saoudien, enfuis sous une grosse pierre puis de venir nous rejoindre au plus vite au 21eme siecle.

merci.

ps : Si vous croisez Doc ou Marty, merci de leurs dire d'arrêter de faire le taxi entre 1985 et 2009.


Et comme disait notre Giscard, euch grand'tringle à'l'glotte qui pendouilloi : 

Au revoir.

cs_cantador · Answer

cantador

Caribensila · Answer

MDR

cs_47 · Answer

Bonsoir,

il existe (non testé donc sans garantie ...), un convertisseur Fortran vers Pascal ici

bonne soirée

Choukri_m · Answer

Merci 47 je vais l'essayé

stevengouti · Answer

il est nul ton code ::: aprend a programmer modulaire : utilise les boucle et les procedures et fonctions pour organiser et d avoir une meilleur complexité

Du fortran au turbo pascal 7

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