Dear linux fortran 77 programers/experts:

I borrow a numerical analysis book from my city's library, (Carnahan 's (applied numerical methods) published 1969 by son john wily), I tried to copy in/test the first chapter, first example, example 1.1
on my gfortran on linux

I found I have difficult to insert input data, after I rund ./a.out

the input data, according to that book, is in page 24

I only can type in first 2 rows, then the program response(for one line of title, that's it, no numerical data as output)

please help, thanks your effort a lot in advance, Eric
/* you might show some link/url/discussing forum about fortran77 and numerical analysis */

You should use an input-file:
So you could quote the input file along with the program and the error message. (Just for those who aren't clairvoyants.)

Script started on Sun 09 Aug 2015 10:38:39 AM PDT
]0;eric@ubuntu: ~/fortranprogramseric@ubuntu:~/fortranprograms$ cat d1_1
8 6
0.0000 0.20000 0.30000 0.40000 0.60000
0.7000 0.9000 1.0000
0.25 1
0.25 2
0.2500 3
0.2500 4
0.2500 5
0.2500 6
0.4500 1
0.4500 2
0.4500 3
0.0500 1
0.0500 2
0.0500 3
0.9500 1
0.9500 2
0.9500 3
0.1000 4
-0.100 4
0.5500 7
1.1000 1
2.000 1
2.0000 1
2.0000 2
2.0000 3
2.0000 4
2.0000 5
2.0000 6
]0;eric@ubuntu: ~/fortranprogramseric@ubuntu:~/fortranprograms$ ./a.out < d1_1
At line 19 of file example1_1.f (unit = 5, file = 'stdin')
Fortran runtime error: Bad value during floating point read
]0;eric@ubuntu: ~/fortranprogramseric@ubuntu:~/fortranprograms$ exit
exit

Script done on Sun 09 Aug 2015 10:39:04 AM PDT
-----------------------------------------------------------------------------------------------------------------------
if I typed in by hand of that d1_1 at ./a.out
again , after 2 line, it will response


/*------------------- */
Script started on Sun 09 Aug 2015 10:42:20 AM PDT
]0;eric@ubuntu: ~/fortranprogramseric@ubuntu:~/fortranprograms$ ./a.out
8 6
0.0 0.2 0.3 0.4 0.6 0.7 0.9 1.0
At line 19 of file example1_1.f (unit = 5, file = 'stdin')
Fortran runtime error: Bad value during floating point read
]0;eric@ubuntu: ~/fortranprogramseric@ubuntu:~/fortranprograms$ ./a.out
8 6
0.0 0.2 0.3 0.4 0.6
1THE SAMPLE FUNCTIONAL VALUES ARE
0 I, X(I) Y(I)

]0;eric@ubuntu: ~/fortranprogramseric@ubuntu:~/fortranprograms$ exit
exit

Script done on Sun 09 Aug 2015 10:43:40 AM PDT

/**-----------------------------------------------------------------------------------**/
so please help,
why its read and its format is hard to correspond my input format?
Eric

Please do something with your post to make it human-readable. Use [code] and [/code] tags for codes/lists.
Also paste the source-program.

Hi fsshl

make sure that you have the OLD legacy gcc3.3( or 3.4)-fortran compiler and tools installed

F77( g77) is not in the gcc 4.? or gcc5
it is in the OLD version 3 compiler

if you ran this
that would have installed the F95 Gfortran-95 software

for the 1977 version
also think of the date on that book

while somethings do not change ... other things DO, like the hardware used in 1969

C Applied Numerical Methods, Example 1.1
C NEWTON'S DIVIDED-DIFFERENCE INTERPOLATING POLYNOMIAL
C
C TEST PROGRAM FOR THE SUBROUTINE DTABLE AND THE FUNCTION
C FNEWT. THIS PROGRAM READS A SET OF N VALUES X(1)...X(N),
C COMPUTE A CORRESPONDING SET OF VALUES Y(1)...Y(N) WHERE
C Y(1)=COS(X(1)), AND THEN CALLS ON SUBROUTINE DTABLE
C TO CALCULATE ALL FINITE DIVIDED DIFFERENCES OF ORDER M OR
C LESS. WITH THE DIVIDED DIFFERENCES STORED IN MATRIX TABLE,
C THE PROGRAM READS VALUES FOR XARG, THE INTERPOLATION
C ARGUMENT, AND IDEG, THE DEGREE OF THE INTERPOLATING
C POLYNOMIAL TO BE EVALUATED BY THE FUNCTION FNEWT.
C FNEWT COMPUTES THE INTERPOLANT VALUE, YINTER, WHICH IS
C COMPARED WITH THE TRUE VALUE, TRUVAL = COS(XARG).
C
DIMENSION X(20), Y(2), TABLE(20, 20)
C
C ..... READ DATA, COMPUTE Y VALUES, AND PRINT .....
READ (5,100) N, M, (X(I), I=1,N)
WRITE (6, 200)
DO 1 I=1,N
Y(I) = COS(X(I))
1 WRITE (6,201) I, X(I), Y(I)
C
C ..... COMPUTE AND PRINT DIVIDED DIFFERENCES .....
CALL DTABLE( X,Y,TABLE,N,M,TRUBL,20 )
IF (TRUBL.NE.0.0) CALL EXIT
WRITE (6,202) M
NM1 = N - 1
DO 6 I=1,NM1
L = 1
IF (I.GT.M) L = M
6 WRITE (6,203) (TABLE(I,J), J=1,L)
C
C ..... READ XARG AND IDEG, CALL ON FNEWT TO INTERPOLATE ....
WRITE (6,204)
7 READ(5,101) XARG, IDEG
YINTER = FNEWT( X,Y,TABLE,N,M,IDEG,XARG,TRUBL,20 )
C
C ..... COMPUTE TRUE VALUE OF COS(XARG) AND PRINT RESULTS ......
TRUVAL = COS(XARG)
WRITE (6,205) XARG, IDEG, YINTER, TRUVAL, TRUBL
GO TO 7
C
C ..... FORMATS FOR INPUT AND OUTPUT STATEMENTS .....
100 FORMAT ( 4X, I3, 10X, I3 / (15X, 5F10.4) )
101 FORMAT ( 7X, F8.4, 13X, I3 )
200 FORMAT ( 33h1THE SAMPLE FUNCTIONAL VALUES ARE / 5H0 I, 8X,
1 4HX(I), 9X, 4HY(I) / 1H )
201 FORMAT ( 1H, I4, 2F13.6 )
202 FORMAT ( 8H1FOR M = , I2, 29H, THE DIVIDED DIFFERENCES ARE )
203 FORMAT ( 1H / (1H, 8E16.7) )
204 FORMAT ( 25H1THE DATA AND RESULTS ARE / 1H0, 5X, 4HXARG, 5X,
1 4HIDEG, 5X, 6HYINTER, 6X, 6HTRUVAL, 3X, 5HTRUBL / 1H )
205 FORMAT ( 1H, F9.4, I8, 2F12.6, F7.1 )
C
END

SUBROUTINE DTABLE ( X,Y,TABLE,N,M,TRUBL,K )
C
C DTABLE COMPUTES THE FINITE DIVIDED DIFFERENCES OF
C Y(1)...Y(N) FOR ALL ORDERS M OR LESS AND STORES THEM IN
C THE LOWER TRIANGULAR PORTION OF THE FIRST M COLUMNS OF THE FIRST
C N-1 ROWS OF THE MATRIX TABLE. FOR INCONSISTENT ARGUMENTS,
C TRUBL = 1.0 ON EXIT. OTHERWISE, TRUBL = 0.0 ON EXIT.
C
DIMENSION X(N), Y(N), TABLE(K, K)
C
C ..... CHECK FOR ARGUMENT CONSISTENCY
IF (M.LT.N) GO TO 2
TRUBL = 1.0
RETURN
C
C ...... CALCULATE FIRST=ORDER DIFFERENCES .....
2 NM1 = N - 1
DO 3 I=1,NM1
3 TABLE(I,1) = (Y(I+1) - Y(I))/(X(I+1) - X(I))
IF (M.LE.1) GO TO 6
C
C ...... CALCULATE HIGHER-ORDER DIFFERENCES .....
DO 5 J=2,M
DO 5 I=J,NM1
ISUB = I+1-J
5 TABLE(I,J) = (TABLE(I,J-1) - TABLE(I-1,J-1))/(X(I+1) - X(ISUB))
C
6 TRUBL = 0.0
RETURN
C
END

FUNCTION FNEWT ( X,Y,TABLE,N,M,IDEG,XARG,TRUBL,K )
C
C FNEWT ASSUMES THAT X(1)...X(N) ARE IN ASCENDING ORDER AND
C FIRST SCANS THE X VECTOR TO DETERMINE WHICH ELEMENT IS
C NEAREST (.GE.) THE INTERPOLATION ARGUMENT, XARG
C THE IDEG+1 BASE POINTS NEEDED FOR THE EVALUATION OF THE
C DIVIDED-DIFFERENCE POLYNOMIAL OF DEGREE IDEG+1 ARE THEN
C CENTERED ABOUT THE CHOSEN ELEMENT WITH THE LARGEST HAVING
C THE SUBSCRIPT MAX. IT IS ASSUMED THAT THE FIRST M DIVIDED
C DIFFERENCES HAVE BEEN COMPUTED BY THE SUBROUTINE
C DTABLE AND ARE ALREADY PRESENT IN THE MATRIX TABLE.
C MAX IS CHECKED TO INSURE THAT ALL REQUIRED BASE POINTS ARE
C AVAILABLE, AND THE INTERPOLANT VALUE IS COMPUTED USING NESTED
C POLYNOMIAL EVALUATION. THE INTERPOLANT IS RETURNED AS
C THE VALUE OF THE FUNCTION. FOR INCONSISTENT ARGUMENTS,
C TRUBL = 1.0 ON EXIT. OTHERWISE, TRUBL = 0.0 ON EXIT
C
DIMENSION X(N), Y(N), TABLE(K,K)
C
C ...... CHECK FOR ARGUMENT INCONSISTENCY ......
IF (IDEG.LE.M) GO TO 2
TRUBL = 1.0
FNEWT = 0.0
RETURN
C
C ....... SEARCH X VECTOR FOR ELEMENT .GE. XARG .....
2 DO 4 I=1,N
4 IF (I.EQ.N .OR. XARG.LE.X(I)) GO TO 5
5 MAX = I + IDEG/2
C
C ......... INSURE THAT ALL REQUIRED DIFFERENCES ARE IN TABLE .........
IF (MAX.LE.IDEG) MAX = IDEG + 1
IF (MAX.GT.N) MAX = N
C
C .... COMPUTE INTERPOLANT VALUE .......
YEST = TABLE(MAX-1, IDEG)
IF (IDEG.LE.1) GO TO 13
IDEGM1 = IDEG - 1
DO 12 I=1,IDEGM1
ISUB1 = MAX - 1
ISUB2 = IDEG - 1
12 YEST = YEST*(XARG - X(ISUB1)) + TABLE(ISUB1-1,ISUB2)
13 ISUB1 = MAX - IDEG
TRUBL = 0.0
FNEWT = YEST*(XARG - X(ISUB1)) + Y(ISUB1)
RETURN
C
END




C /************************************************************************************************/
this is from that book's example 1.1, 's fortran program, I just copied it, and tested it. if did not work on either gfortran or fort77
but, if I modify that program 's first format for 5f to 8f, and by hand enter at command prompt of ./a.out, I have a chance to survive to
enter 4th line of data,
0.25 1
when I compile my executable code through gfortran

I wish any expert can dig something more useful for me at next post, thanks a lot in advance., eric

Please do something with your post to make it human-readable. Use [code] and [/code] tags for codes/lists.

I modify a little bit typo and using hand input data, I found X(I) and Y(I) are correct(same as book's print), but
table(the divided differences are), are different from book's(and I believe book's is correct, mine is wrong), especially at very begining stage,i.e. TABLE(1,1)
I got it as 0.3295133E+00
(book's is -0.9966671E-01 )

plz help

/* I certain hope to learn the way to input data by file, ./a.out < Data, rathern than type in by hand each time it execute./a.out or ./a.exe; hope some expert give advice */

/* if you are not reading that (same) book, please don't post reply */

1 members found this post helpful.

> if you are not reading that (same) book, please don't post reply

Okay. Also I don't reply because you didn't add the [code] and [/code] tags. Good bye.

After I modify 2 typo(s) (one is Y(2) should change to Y(20) , the other is L = 1 , should change to L = I), I tested it again,
most of my result correspond with books's result, but still not 100 (at high degree of TABLE and (THE DATA AND RESULT)(DEGE =1 , is OK, but after that , DEGE = 2 to 6, all deviate from book's result)
Here is my typescript
---------------------------------------------------------------------------
Script started on Tue, Aug 11, 2015 9:35:24 AM
]0;~/fortranprograms Erli89080@Erli89080-PC ~/fortranprograms $ a../a.exe 
N = 8 M = 6
X(1)...X(5) = 0.0000 0.2000 0.3000 0.4000 0.6000
X(6)...X(8) = 0.7000 0.9000 1.0000
1THE SAMPLE FUNCTIONAL VALUES ARE
0 I, X(I) Y(I)

, 1 0.000000 1.000000
, 2 0.200000 0.980067
, 3 0.300000 0.955337
, 4 0.400000 0.921061
, 5 0.600000 0.825336
, 6 0.700000 0.764842
, 7 0.900000 0.621610
, 8 1.000000 0.540302
1.00000000
0.980066597
-1.99334025E-02
1FOR M = 6, THE DIVIDED DIFFERENCES ARE

, -0.9966701E-01

, -0.2473008E+00 -0.4921128E+00

, -0.3427553E+00 -0.4772725E+00 0.3710069E-01

, -0.4786268E+00 -0.4529047E+00 0.6091952E-01 0.3969805E-01

, -0.6049349E+00 -0.4210273E+00 0.7969350E-01 0.3754797E-01 -0.3071550E-02

, -0.7161609E+00 -0.3707533E+00 0.1005480E+00 0.3475756E-01 -0.3986300E-02 -0.1016389E-02

, -0.8130769E+00 -0.3230534E+00 0.1192497E+00 0.3116940E-01 -0.5125936E-02 -0.1424545E-02
1THE DATA AND RESULTS ARE
0 XARG IDEG YINTER TRUVAL TRUBL

XARG = 2.0000 IDEG = 1
, 2.0000 1 -0.272775 -0.416147 0.0
XARG = 2.0000 IDEG = 2
, 2.0000 2 -0.628133 -0.416147 0.0
XARG = 2.0000 IDEG = 3
, 2.0000 3 -0.062670 -0.416147 0.0
XARG = 2.0000 IDEG = 4
, 2.0000 4 1.519942 -0.416147 0.0
XARG = 2.0000 IDEG = 5
, 2.0000 5 1.216805 -0.416147 0.0
XARG = 2.0000 IDEG = 6
, 2.0000 6 0.932131 -0.416147 0.0

]0;~/fortranprograms Erli89080@Erli89080-PC ~/fortranprograms $ exit
exit

Script done on Tue, Aug 11, 2015 9:45:27 AM
------------------------------------------------------------------------------------------------------------------------
I am on cygwin(64 bit), compile by gfortran, on my pc with intel core 2 duo processor, in window vista ultrimate 64 bits
(book author said he did it in IBM360/370)

plz help

I found my final typo(s), and fix them.
Now it work 100 corresponding with book's result.
thanks your concern and help /* it can be small homework for novice */