通过派生类型数组 [英] Pass derived type as array

问题描述

在Fortran语言，可以在阵列上运行，但一个人如何可以把一个派生类型的指数作为数组太大的一部分？ code会解释什么，我想要做的最好的：

In Fortran, one can operate on arrays, but how can one treat the indices of a derived type as part of an array too? Code would explain what I want to do best:

type mytype
    integer :: b(3,3)
    real :: c(4)
endtype

integer :: a(3,3)
real :: d(2,4)
type(mytype) :: mat(2)

!do stuff so that 'mat' gets values
....

!usually one does this
a = matmul(mat(1)%b, transpose(mat(2)%b))

!multiplying two 3x3 matrices

!but how does one do this? Note the "array"
d = matmul(mat(:)%c, mat(:)%c)

我假定最后一行类似于一个2x4的矩阵被乘以本身。然而，当我尝试编译，gfortran抱怨

I assumed that the final line is analogous to a 2x4 matrix being multiplied with itself. However, when I try to compile, gfortran complains

错误：有非零秩的两个或多个部分引用，不得指定

Error: Two or more part references with nonzero rank must not be specified

这是可能用Fortran办？

Is this possible to do in Fortran?

推荐答案

您希望编译器把垫（：）％C 为2×4矩阵？它不这样的。 垫和 C 是不同的对象和他们的等级不合并成一个单一的阵列。 垫是一个用户定义的类型和 C 是一个真正的矩阵。仅仅因为你只使用 C 的-component 垫并不意味着编译器会推动<$ C $ ç> C 到一个更高维实阵列，根据垫。

You want the compiler to regard mat(:)%c as a 2 x 4 matrix? It doesn't work that way. mat and c are different objects and their ranks don't merge into a single array. mat is a user-defined type and c is a real matrix. Just because you are only using the c-component of mat doesn't mean the compiler will promote c to a higher dimensional real array, based on the dimension of mat.

您可以通过创建一个新的数组 X = [垫（1）％C，垫（2）％C] 。你可以使用重塑来控制形状。

You could create a new array via X = [ mat(1)%c, mat(2)%c ]. You could use reshape to control the shape.

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