在Fortran中将矩阵用作函数中的参数和子例程中的输出
在Fortran中将矩阵用作函数中的参数和子例程中的输出
提问于 2021-02-16 03:27:59
我试图创建一个程序，它要求我使用矩阵作为函数和子例程的输入，并要求我在Fortran中将矩阵作为子例程输出。但是，我在这样做的时候遇到了多个错误。我不能理解这些错误的来源以及如何修复它们。我对逻辑很有信心，但我似乎在处理矩阵时犯了错误。
解线性方程组的程序(部分旋转高斯消去法)
代码：
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错误：
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谢谢!
Stack Overflow用户
发布于 2021-02-16 07:04:14
这是一个考虑到@SteveLionel的建议和以下注释的工作示例：
- 始终使用
implicit none，至少在主程序中使用一次，并且不要忘记将-warn标志传递给compiler. - Either使用函数和子例程的代码，然后将
use <module>添加到主程序中，或者像我下面所做的那样，简单地使用contains并将它们包含在主程序中。 -
interchanged数组已经在主程序中分配，您不需要在interchanger子例程中重新分配它，只需将其作为假定形式的code未使用变量传递即可；为了可读性，最好将alpha, maximum, k, inte. - Define function.
- Function类型的interface.
- The
a写在函数名前面；除非您正在使用显式的f_maxfinderf_maxfinder过程接受real输入，否则请参阅f_maxfinder的定义，不要在main程序中再次声明该函数，您不需要它并在function/subroutine.
中添加任何缺少的变量声明
program solving_equations
implicit none
real, allocatable :: a(:,:), interchanged(:,:), x(:)
real :: addition, multiplying_term
integer :: i, j, row, rth_ele, n, s
read (*,*) n
allocate ( a( n,(n+1) ) )
allocate ( x( n ) )
allocate ( interchanged( n,(n+1) ) )
do i = 1,n
do j = 1,(n+1)
read (*,*) a(i,j)
end do
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
contains
integer function f_maxfinder(a, n, rth_ele)
integer :: n, rth_ele, inte
real :: maximum, a(:,:)
maximum = a(rth_ele,rth_ele)
do inte = n,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, rth_ele, row, n
real :: alpha, a(:,:), interchanged(:,:)
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
end program solving_equations输入一个示例3x4数组，您将获得以下输出(检查结果，您知道您的算法)：
3
4
3
6
3
7
4
6
7
4
4
2
0
2.05263186
-2.15789509
0.210526198
Process returned 0 (0x0) execution time : 1.051 s
Press any key to continue.页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/66214235
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