[求助]fortran这种用法百思不得其解

airsupply

他的x,y,z的维数是x(jdim,kdim,idim),y(jdim,kdim,idim),z(jdim,kdim,idim)
却用到了
      t(izz,1) = x(izz+1,1,i)-x(izz,2,i)
      t(izz,2) = y(izz+1,1,i)-y(izz,2,i)
      t(izz,3) = z(izz+1,1,i)-z(izz,2,i)
      t(izz,4) = x(izz+1,2,i)-x(izz,1,i)
      t(izz,5) = y(izz+1,2,i)-y(izz,1,i)
      t(izz,6) = z(izz+1,2,i)-z(izz,1,i)
其中izz最大值居然到了jdim*kdim-jdim，明显会数组越界，visual fortran编译通过了，运行的时候报错数组越界。但用linux下的intel fortran编译器编译通过计算了，实在是搞不懂。
附上下面一段程序：

      parameter(nn=maxbl)
c
      character*80 bcfilei,bcfilej,
     .             bcfilek
c
      dimension x(jdim,kdim,idim),y(jdim,kdim,idim),z(jdim,kdim,idim)
      dimension sj(jdim*kdim,idim-1,5),sk(jdim*kdim,idim-1,5),
     .          si(jdim*kdim,idim,5)
      dimension t(jdim*kdim,6),t1(jdim*kdim,idim,5)
c
      common /sklton/ isklton
      common /bcnew/ bcvali(nn,maxseg,5,2),
     .        bcvalj(nn,maxseg,5,2),bcvalk(nn,maxseg,5,2),
     .        nbci0(nn),nbcidim(nn),nbcj0(nn),nbcjdim(nn),
     .        nbck0(nn),nbckdim(nn),ibcinfo(nn,maxseg,7,2),
     .        jbcinfo(nn,maxseg,7,2),kbcinfo(nn,maxseg,7,2)
      common /bcnew2/ bcfilei(nn,maxseg,2),bcfilej(nn,maxseg,2),
     .        bcfilek(nn,maxseg,2)
      common /twod/ i2d
      common /zero/iexp
c
c***********************************************************************
c  metrics:
c          si(1-3)...components of unit normal to i-face (directed area)
c          si( 4 )...area of  i-face
c          si( 5 ).. speed of i-face
c
c          sj(1-3)...components of unit normal to j-face (directed area)
c          sj( 4 )...area of  j-face
c          sj( 5 ).. speed of j-face
c
c          sk(1-3)...components of unit normal to k-face (directed area)
c          sk( 4 )...area of  k-face
c          sk( 5 ).. speed of k-face
c
c  notes:
c        1) the normal to a cell face is determined via the cross product
c           of two diagonal vectors connecting oposite corners of the cell
c           face.
c        2) a unit normal is obtained by dividing by the magnitude
c           of the inner product.
c        3) The area of the cell face is given by  one-half the magnitude
c           of the cross product of the two diagonal vectors
c        4) in a jdim*kdim*idim grid there are really only
c                   idim*(jdim-1)*(kdim-1) i-faces
c                   jdim*(idim-1)*(kdim-1) j-faces
c                   kdim*(idim-1)*(jdim-1) k-faces
c           however the metric arrays are dimensioned larger than this -
c           for efficient vectorization, loops throughout the code run
c           over these fictitious faces. thus, for safety, the metrics
c           for these fictitious faces are set to the corresponding metrics
c           for the neighboring face.
c
c        5) unsteady metric terms si(5), sj(5), sk(5) are set to zero
c           here, and must be reset elsewhere if the mesh is actually
c           in motion
c
c        6) tolerance to automatically detect collapsed (singular) metrics
c           is set by atol
c
c***********************************************************************
c
c     tolerance for collapsed metrics (10.**(-iexp) is machine zero)
c
      atol = max(1.e-07,10.**(-iexp+1))
c
      jdim1 = jdim-1
      kdim1 = kdim-1
      idim1 = idim-1
c
c***********************************************************************
c
c     metrics for i=constant surfaces
c
c***********************************************************************
c
c     *** interior faces ***
c
      n = jdim*kdim-jdim
   
c    do 8211  izz=1,n
c    write(8211,*) x(izz+1,1,1)
c8211    continue
c    continue
   
      
c
      do 1040 i=2,idim1
cdir$ ivdep
      do 1050 izz=1,n
c
c     components of vectors connecting opposite corners of cell j,k
      t(izz,1) = x(izz+1,1,i)-x(izz,2,i)
      t(izz,2) = y(izz+1,1,i)-y(izz,2,i)
      t(izz,3) = z(izz+1,1,i)-z(izz,2,i)
      t(izz,4) = x(izz+1,2,i)-x(izz,1,i)
      t(izz,5) = y(izz+1,2,i)-y(izz,1,i)
      t(izz,6) = z(izz+1,2,i)-z(izz,1,i)
c
c     cross product of vectors
      si(izz,i,1) =  t(izz,2)*t(izz,6)-t(izz,3)*t(izz,5)
      si(izz,i,2) = -t(izz,1)*t(izz,6)+t(izz,3)*t(izz,4)
      si(izz,i,3) =  t(izz,1)*t(izz,5)-t(izz,2)*t(izz,4)
c
c     magnitude of cross product
      si(izz,i,4) = si(izz,i,1)*si(izz,i,1)+si(izz,i,2)*si(izz,i,2)+
     .              si(izz,i,3)*si(izz,i,3)
1050 continue

lychu326

  这个问题很好解释,通常Fortran的数组是按列存储的.
  x(i,j,k)的地址计算是  i-1+(j-1)*imax+(k-1)*imax*jmax
但如果编译器中的数组是按行存储(C语言方式)
  其地址计算是 (i-1)*jmax*kmax+(j-1)*kmax+k-1.
判断是否越界,是看数组的地址计算方式超出了范围