topic Code resulting infinity values in Intel® Fortran Compiler

Code resulting infinity values

Henry26 — Thu, 20 Oct 2022 18:08:41 GMT

Hello, everyone.

Please, could somebody help-me with the code below? It is resulting in infinity value.

PROGRAM MR !============================================================ !DECLARACAO DAS VARIAVEIS !============================================================ IMPLICIT NONE REAL, EXTERNAL::F INTEGER, PARAMETER :: N=6, NPC=4 INTEGER I, J REAL :: X(0:N),Y(0:N),A(0:6),B(0:6),HP,XP(0:NPC),P(0:NPC) OPEN(UNIT=20, FILE="FUNCA.txt", STATUS="UNKNOWN") !============================================================ !DADOS DA FUNCAO INTERPOLADA !============================================================ PRINT*, "DADOS DA FUNCAO" DO I=0, N X=X(I)+1 Y(I)=F(X) WRITE(20,*)X(I),Y(I) WRITE(*,*)I,X(I),Y(I) END DO !============================================================ !COEFICIENTES DAS FUNCOES INTERPOLADAS !============================================================ PRINT*, "COEFICIENTE DAS FUNCOES" DO I=0, N-1 A(I)=(Y(I+1)-Y(I))/(X(I+1)-X(I)) B(I)=(Y(I)*X(I+1)-(I)*Y(I+1))/(X(I+1)-X(I)) WRITE (*,*)I,A(I),B(I) END DO END PROGRAM MR !============================================================ !FUNCAO !============================================================ REAL FUNCTION F(X) IMPLICIT NONE REAL*4::x F = SIN(X/2) END

The results is:

DADOS DA FUNCAO 0 1.00000000 0.479425550 1 2.00000000 0.841470957 2 3.00000000 0.997494996 3 4.00000000 0.909297407 4 5.00000000 0.598472118 5 6.00000000 0.141120002 6 7.00000000 -0.350783229 COEFICIENTE DAS FUNCOES 0 Infinity Infinity 1 Infinity Infinity 2 -Infinity Infinity 3 -Infinity Infinity 4 -Infinity Infinity 5 -Infinity Infinity

Thank you!!

Re: Code resulting infinity values

JohnNichols — Thu, 20 Oct 2022 18:38:35 GMT

You need to fix this error first

I am usign the latest compiler IFX and IFORT both complain.

Re: Code resulting infinity values

jimdempseyatthecove — Thu, 20 Oct 2022 18:59:57 GMT

PRINT*, "DADOS DA FUNCAO" DO I=0, N X=X(I)+1

Two issues with the above:

1) At the point of execution of line 3 (line 17 of posted code), the array X was(is) undefined.

2) The statement X=X(I)+1, takes the scalar variable result of the expression "X(I)+1", and assigns it to the entire array X(0:N). In this case X(0:N) receives "undefined value at X(0) +1", and repeats this for "X(1)+1", ... "X(N)+1". And at the end of the loop, "X(N)+1". Note, the entire array of X changes at each iteration (each iteration using undefined values including values defined using undefined values).

EDIT: Strike (1) the above, apparently the code snip above is missing lines that defined the values of X(0:9).

Point (2) may still be of importance.

Jim Dempsey

Re: Code resulting infinity values

Henry26 — Thu, 20 Oct 2022 18:57:32 GMT

Thanks you for the answer @jimdempseyatthecove

In this case, how could I fix it? I just need to make an array from 0-N and after use the values of this array to input in a function

Re: Code resulting infinity values

jimdempseyatthecove — Thu, 20 Oct 2022 19:11:30 GMT

Would the following represent what you intend?

DO I=0, N !**change** X=X(I)+1 X(I)=I+1 !**change** Y(I)=F(X) Y(I)=F(X(I)) WRITE(20,*)X(I),Y(I) WRITE(*,*)I,X(I),Y(I) END DO

Re: Code resulting infinity values

jimdempseyatthecove — Thu, 20 Oct 2022 19:15:36 GMT

Note, the second change is what JohnNichols pointed out as an error to line 18,

X is a rank-1 array

F(X) is a rank-1 array of SIN's of the array X.

This cannot be assigned to a scalar.

Jim Dempsey

Re: Code resulting infinity values

Henry26 — Thu, 20 Oct 2022 19:36:48 GMT

Now it works!! Thank you so much!!!