File: setlvl.f

package info (click to toggle)
nastran 0.1.95-2
  • links: PTS, VCS
  • area: non-free
  • in suites: bookworm, bullseye, sid
  • size: 122,540 kB
  • sloc: fortran: 284,409; sh: 771; makefile: 324
file content (119 lines) | stat: -rw-r--r-- 4,035 bytes parent folder | download | duplicates (2)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
      SUBROUTINE SETLVL (NEWNM,NUMB,OLDNMS,ITEST,IBIT)
C
C     CREATES A NEW SUBSTRUCTURE NEWNM WHERE
C     - NEWNM IS AN INDEPENDENT SUBSTRUCTURE IF NUMB = 0
C     - NEWNM IS REDUCED FROM THE FIRST SUBSTRUCTURE IN THE ARRAY OLDNMS
C     - NEWNM RESULTS FROM COMBINING THE FIRST I SUBSTRUCTURES IN THE
C       ARRAY OLDNMS IF NUMB = I
C
C     THE OUTPUT VARIABLE ITEST TAKES ON ONE OF THE FOLLOWING VALUES
C          4  IF ONE  OR MORE SUBSTRUCTURES IN OLDNMS DO NOT EXIST
C          7  IF NEWNM ALREADY EXISTS
C          8  IF ONE OF THE SUBSTRUCTURES IN OLDNMS HAS ALREADY
C             BEEN USED IN A REDUCTION OR COMBINATION
C          1  OTHERWISE
C
C     IF ITEST IS SET TO 4, NUMB WILL BE SET TO THE NUMBER OF
C     SUBSTRUCTURES IN OLDNMS THAT DO NOT EXIST AND THE FIRST NUMB NAMES
C     IN OLDNMS WILL BE SET TO THE NAMES OF THOSE SUBSTRUCTURES THAT DO
C     NOT EXIST.  BIT IBIT OF THE FIRST MDI WORD IS SET TO INDICATE THE
C     APPROPRIATE TYPE OF SUBSTRUCTURE. IF IBIT IS ZERO NO CHANGE IS
C     MADE TO THE MDI
C
      IMPLICIT INTEGER (A-Z)
      EXTERNAL        LSHIFT,ANDF,ORF,COMPLF
      LOGICAL         DITUP,MDIUP
      DIMENSION       NEWNM(2),OLDNMS(14),IOLD(7),NMSBR(2)
      COMMON /ZZZZZZ/ BUF(1)
      COMMON /SOF   / DIT,DITPBN,DITLBN,DITSIZ,DITNSB,DITBL,
     1                IODUM(8),MDI,MDIPBN,MDILBN,MDIBL,
     2                NXTDUM(15),DITUP,MDIUP
      DATA    IEMPTY/ 4H    /, NMSBR / 4HSETL,4HVL  /
      DATA    LL,CS , HL    /  2,2,2 /
      DATA    IB    / 1     /
C
      CALL CHKOPN (NMSBR(1))
      ITEST = 1
      CALL FDSUB (NEWNM(1),I)
      IF (I  .NE.  -1) GO TO 500
      IF (NUMB .EQ. 0) GO TO 20
C
C     MAKE SURE THAT ALL THE SUBSTRUCTURES IN OLDNMS DO EXIST.
C
      ICOUNT = 0
      DO 10 I = 1,NUMB
      K = 2*(I-1) + 1
      CALL FDSUB (OLDNMS(K),IOLD(I))
      IF (IOLD(I) .GT. 0) GO TO 10
      ICOUNT = ICOUNT + 1
      KK = 2*(ICOUNT-1) + 1
      OLDNMS(KK  ) = OLDNMS(K  )
      OLDNMS(KK+1) = OLDNMS(K+1)
   10 CONTINUE
      IF (ICOUNT .EQ. 0) GO TO 20
      NUMB = ICOUNT
      GO TO 510
   20 CALL CRSUB (NEWNM(1),INEW)
      IF (NUMB .EQ. 0) RETURN
C
C     NEWNM IS NOT A BASIC SUBSTRUCTURE (LEVEL 0).
C     UPDATE NEWNM S DIRECTORY IN THE MDI.
C
      CALL FMDI (INEW,IMDI)
      LLMASK = COMPLF(LSHIFT(1023,20))
      BUF(IMDI+LL) = ORF(ANDF(BUF(IMDI+LL),LLMASK),LSHIFT(IOLD(1),20))
      IF (IBIT .NE. 0) BUF(IMDI+IB) = ORF(BUF(IMDI+IB),LSHIFT(1,IBIT))
      MDIUP = .TRUE.
C
C     UPDATE IN THE MDI THE DIRECTORIES OF THE SUBSTRUCTURES IN OLDNMS.
C
      IF (NUMB .GT. 7) NUMB = 7
      MASKCS = COMPLF(LSHIFT(1023,10))
      DO 50 I = 1,NUMB
      CALL FMDI (IOLD(I),IMDI)
      IF (ANDF(BUF(IMDI+HL),1023) .EQ. 0) GO TO 40
      ICOUNT = I
      GO TO 520
   40 BUF(IMDI+HL) = ORF(BUF(IMDI+HL),INEW)
      MDIUP = .TRUE.
      IF (NUMB .EQ. 1) RETURN
      IF (I .EQ. NUMB) GO TO 130
      BUF(IMDI+CS) = ORF(ANDF(BUF(IMDI+CS),MASKCS),LSHIFT(IOLD(I+1),10))
   50 CONTINUE
  130 BUF(IMDI+CS) = ORF(ANDF(BUF(IMDI+CS),MASKCS),LSHIFT(IOLD(1),10))
      RETURN
C
C     NEWNM ALREADY EXISTS.
C
  500 ITEST = 7
      RETURN
C
C     ONE OR MORE OF THE SUBSTRUCTURES IN OLDNMS DO NOT EXIST.
C
  510 ITEST = 4
      RETURN
C
C     ONE OF THE SUBSTRUCTURES IN OLDNMS HAS ALREADY BEEN USED IN A
C     REDUCTION OR COMBINATION.  REMOVE ALL CHANGES THAT HAVE BEEN MADE.
C
  520 ITEST = 8
      CALL FDIT (INEW,IDIT)
      BUF(IDIT  ) = IEMPTY
      BUF(IDIT+1) = IEMPTY
      DITUP = .TRUE.
      IF (2*INEW .NE. DITSIZ) GO TO 525
      DITSIZ = DITSIZ - 2
  525 DITNSB = DITNSB - 1
      CALL FMDI (INEW,IMDI)
      BUF(IMDI+LL) = ANDF(BUF(IMDI+LL),LLMASK)
      MDIUP  = .TRUE.
      ICOUNT = ICOUNT - 1
      IF (ICOUNT .LT. 1) RETURN
      DO 530 I = 1,ICOUNT
      CALL FMDI (IOLD(I),IMDI)
      BUF(IMDI+HL) = ANDF(BUF(IMDI+HL),COMPLF(1023))
      BUF(IMDI+CS) = ANDF(BUF(IMDI+CS),MASKCS)
      MDIUP = .TRUE.
  530 CONTINUE
      RETURN
      END