在 Fortran 中使用矩阵作为函数中的参数和子例程中的输出

我试图创建一个程序，要求我使用矩阵作为函数和子例程的输入，并且还要求我将矩阵作为 Fortran 中的子例程输出。但是，我在这样做时遇到了多个错误。我无法理解这些错误的根源以及如何修复它们。 我对逻辑很有信心，但我在处理矩阵时似乎犯了错误。

求解线性方程组的程序（部分旋转的高斯消元法）

Code:

program solving_equations
implicit none

real, allocatable :: a(:,:),interchanged(:,:)
real, allocatable :: x(:)
real addition,multiplying_term,alpha,maximum
integer i,j,row,rth_ele,f_maxfinder,k,n,s,inte

read(*,*)n
allocate(  a( n,(n+1) )  )
allocate( x(n) )
allocate(  interchanged( n,(n+1) )  )

do i=1,n
read(*,*)( a(i,j),j=1,(n+1) )
end do

do rth_ele= 1,(n-1) 
  row=f_maxfinder( a , n , rth_ele )

  if (row==rth_ele) then
  continue

  else
 call interchanger(a,rth_ele,row,n,interchanged)
 a = interchanged
  
  end if      
        
        do i= (rth_ele+1) , n
        ! once i is fixed, multiplying term is fixed too
        multiplying_term=( a(i,rth_ele)/a(rth_ele,rth_ele) )
            
            do j=1,(n+1)
            a(i,j)=a(i,j)-a(rth_ele,j)*multiplying_term
            end do
       end do   
end do      

x(n)=a(n,n+1)/a(n,n)
do i=(n-1),1,-1
addition=0.0
    
    do s=n , (i+1) , -1
    addition=addition+a(i,s)*x(s)
    end do

x(i)= (   ( a(i,n+1)- addition  )/a(i,i)  )
end do

do i=1,n
print*,x(i)
end do

endprogram solving_equations

!=================

function f_maxfinder(a,n,rth_ele)
integer inte,f_maxfinder
real maximum
maximum=a(rth_ele,rth_ele)

do inte=n,nint(rth_ele+1),-1
  if(  a(inte,rth_ele) > maximum  ) then
  maximum = a(inte,rth_ele)
  f_maxfinder=inte
  
  else
  continue
  
  end if
end do
end  


subroutine interchanger( a,rth_ele,row,n,interchanged )
integer i
real alpha
real, allocatable :: interchanged(:,:)
allocate(  interchanged( n,(n+1) )  )

do i=1,n+1
alpha=a(row,i)
a(row,i)=a(rth_ele,i)
a(rth_ele,i)=alpha
end do

do i=1,n
do j=1,(n+1)
interchanged(i,j)=a(i,j)
end do
end do

end

Errors:

   row=f_maxfinder( a , n , rth_ele )
                   1
Warning: Rank mismatch in argument 'a' at (1) (scalar and rank-2)

 a(row,i)=a(rth_ele,i)

Error: The function result on the lhs of the assignment at (1) must have the pointer attribute.

 a(rth_ele,i)=alpha

Error: The function result on the lhs of the assignment at (1) must have the pointer attribute.

  call interchanger(a,rth_ele,row,n,interchanged)
                  1
Error: Explicit interface required for 'interchanger' at (1): allocatable argument

Thanks!

1. 你缺少一个声明a作为 f_maxfinder 中的数组。implicit none是你的朋友——一定要一直使用它。
2. interchanger有一个虚拟参数interchanged这是一个可分配的、假定形状的数组。这需要一个显式接口interchanger在调用者中可见。 （见我的帖子https://stevelionel.com/drfortran/2012/01/05/doctor-fortran-gets-explicit-again/ https://stevelionel.com/drfortran/2012/01/05/doctor-fortran-gets-explicit-again/了解更多相关信息。

可以通过将子例程放入模块并添加一个来解决接口问题use主程序中的模块。

顺便说一句，没有必要做a可分配于f_maxfinder，因为您没有分配或取消分配它。它仍然是一个假定形状的数组，因此仍然需要显式接口。