Fortran对矩阵的赋值报错于输出

sqs

对于6*6或者更高阶的矩阵（二维），怎么简单的对里面的每个元素进行赋值呢（包括我的矩阵当中存在较多的元素为零）?
另外这个报错是为什么，怎么修改呢？
另外用print输出矩阵时，如何分行呢？尤其我矩阵当中的元素还是复数，write输出矩阵的格式是什么呢？

[Fortran] 纯文本查看 复制代码

Program workfile
        Implicit none
        integer , parameter :: n = 1000
        real , parameter :: t1 = 3.12 , t2 = 0.29 , t3 = -0.0103
        complex , parameter :: i = (0, 1) , t0 = (0,0)
        real , parameter :: PI = acos(-1.0) , a0 = 1.42e-10
        real :: b1(2), b2(2) , a1(2) , a2(2) , m1 , m2
        complex H(3, 2)
        real , allocatable :: k(:,:,:)
        complex :: f1 , f2
        integer :: x , y
        allocate(k(2,n-1,n-1))
        b1 = [ 2*PI/a0*sqrt(3.0) , 2*PI/a0 ]
        b2 = [ 2*PI/a0*sqrt(3.0) , -2*PI/a0 ]
        a1 = [ a0*sqrt(3.0)/2 , a0/2 ]
        a2 = [ a0*sqrt(3.0)/2 , -a0/2 ]
        Forall(x=1:size(k,2), y=1:size(k,3))
                k(:,x,y) = (b1*x+b2*y)/n
        End Forall
        m1 = dot_product(a1 , k(:,1,2))
        m2 = dot_product(a2 , k(:,1,2))
        f1 = 1+exp(i*m1)+exp(i*m2)
        f2 = 1+exp(-i*m1)+exp(-i*m2)
        H = reshape((/t0, t1*f1, t0, t0, t0, t0,&
                t1*f2, t0, t0, t2, t0, t3,&
                t0, t0, t0, t1*f2, t0 ,t0,&
                t0, t2, t1*f1,t0, t0, t2,&
                t0, t0, t0, t0, t0, t1*f1,&
                t0, t3, t0, t0, t1*f2, t0/),(/6 , 6/)
        print * , H(:,:)
        End Program workfile

报错如下：

[Fortran] 纯文本查看 复制代码

work.f90:25:30:

   25 |                 t1*f2, t0, t0, t2, t0, t3,&
      |                              1
Error: Element in COMPLEX(4) array constructor at (1) is REAL(4)

file:///C:\Users\Lenovo\AppData\Roaming\Tencent\Users\2597505881\QQ\WinTemp\RichOle\]U}@9S9_7O64~BG4H02K8VH.png

vvt

[Fortran] 纯文本查看 复制代码

Program workfile
  Implicit none
  integer , parameter :: n  = 1000
  real    , parameter :: t1 = 3.12 
  complex , parameter :: i = (0, 1) , t0 = (0,0), t2 = 0.29 , t3 = -0.0103
  real , parameter :: PI = acos(-1.0) , a0 = 1.42e-10
  real :: b1(2), b2(2) , a1(2) , a2(2) , m1 , m2
  complex H(6, 6)
  real , allocatable :: k(:,:,:)
  complex :: f1 , f2
  integer :: x , y
  allocate(k(2,n-1,n-1))
  b1 = [ 2*PI/a0*sqrt(3.0) , 2*PI/a0 ]
  b2 = [ 2*PI/a0*sqrt(3.0) , -2*PI/a0 ]
  a1 = [ a0*sqrt(3.0)/2 , a0/2 ]
  a2 = [ a0*sqrt(3.0)/2 , -a0/2 ]
  Forall(x=1:size(k,2), y=1:size(k,3))
    k(:,x,y) = (b1*x+b2*y)/n
  End Forall
  m1 = dot_product(a1 , k(:,1,2))
  m2 = dot_product(a2 , k(:,1,2))
  f1 = 1+exp(i*m1)+exp(i*m2)
  f2 = 1+exp(-i*m1)+exp(-i*m2)
  H = reshape([t0, t1*f1, t0, t0, t0, t0,&
    t1*f2, t0, t0, t2, t0, t3,&
    t0, t0, t0, t1*f2, t0 ,t0,&
    t0, t2, t1*f1,t0, t0, t2,&
    t0, t0, t0, t0, t0, t1*f1,&
    t0, t3, t0, t0, t1*f2, t0],shape(H))
  write(*,"(*(6('(',f8.5,',',f8.5,')',1x),/) )") H(:,:)
End Program workfile

Transpose

[Fortran] 纯文本查看 复制代码

Program workfile
    Implicit none
    integer , parameter :: n = 1000
    real , parameter :: t1 = 3.12 , t2 = 0.29 , t3 = -0.0103
    complex , parameter :: i = (0, 1) , t0 = (0,0)
    real , parameter :: PI = acos(-1.0) , a0 = 1.42e-10
    real :: b1(2), b2(2) , a1(2) , a2(2) , m1 , m2
    complex H(6, 6)
    real , allocatable :: k(:,:,:)
    complex :: f1 , f2
    integer :: x , y
    integer::i1
    allocate(k(2,n-1,n-1))
    b1 = [ 2*PI/a0*sqrt(3.0) , 2*PI/a0 ]
    b2 = [ 2*PI/a0*sqrt(3.0) , -2*PI/a0 ]
    a1 = [ a0*sqrt(3.0)/2 , a0/2 ]
    a2 = [ a0*sqrt(3.0)/2 , -a0/2 ]
    Forall(x=1:size(k,2), y=1:size(k,3))
            k(:,x,y) = (b1*x+b2*y)/n
    End Forall
    m1 = dot_product(a1 , k(:,1,2))
    m2 = dot_product(a2 , k(:,1,2))
    f1 = 1+exp(i*m1)+exp(i*m2)
    f2 = 1+exp(-i*m1)+exp(-i*m2)
    H = reshape([complex::&
            t0, t1*f1, t0, t0, t0, t0,&
            t1*f2, t0, t0, t2, t0, t3,&
            t0, t0, t0, t1*f2, t0 ,t0,&
            t0, t2, t1*f1,t0, t0, t2, &
            t0, t0, t0, t0, t0, t1*f1,&
            t0, t3, t0, t0, t1*f2, t0],[6 , 6])
    do i1=1,6
        write(*,*)H(i1,:)
    end do
End Program workfile

数组构造器里面的数据类型要求相同，t2，t3都是real类型的，所以没法再一起使用，可以给构造器加上 `[complex::]`修饰，这样类型会自动转化为complex类型

输出分行的话，建议用循环，Fortran没有内置复数的描述符。

sqs

Transpose 发表于 2022-3-7 15:32
[mw_shl_code=fortran,true]Program workfile
    Implicit none
    integer , parameter :: n = 1000

好的，感谢！