Fortran 如何返回数组？

发布于 2024-09-15 10:06:47 字数 774 浏览 5 评论 0原文

Fortran 库 CUBPACK 中的子例程 Rule_Tn 需要一个描述积分向量函数的参数 Integrand。因为

INTERFACE 
   FUNCTION Integrand(NF,X) RESULT(Value)
      USE Precision_Model
      INTEGER,                       INTENT(IN)  :: NF
      REAL(KIND=STND), DIMENSION(:), INTENT(IN)  :: X
      REAL(KIND=STND), DIMENSION(NF)             :: Value
   END FUNCTION Integrand
END INTERFACE

我想从 C 代码调用 Rule_Tn，所以我需要在 C 中定义一个与上面的接口完全匹配的函数类型。因此我尝试弄清楚 Fortran 函数如何返回数组。起初我认为以下 C 签名

void Integrand(double* value, const int* nf, const int* x);

与上面的接口匹配。但大错特错了！我遇到了段错误。我已经测试过 double 是 REAL(KIND=STND) 对应的类型，STND 来自模块 Precision_Model。

现在谁能告诉我什么是正确的签名？我正在使用 GNU Fortran 和 C 编译器。

原文

The subroutine Rule_Tn in the Fortran library CUBPACK needs a parameter Integrand describing the integrated vector function. It's a

INTERFACE 
   FUNCTION Integrand(NF,X) RESULT(Value)
      USE Precision_Model
      INTEGER,                       INTENT(IN)  :: NF
      REAL(KIND=STND), DIMENSION(:), INTENT(IN)  :: X
      REAL(KIND=STND), DIMENSION(NF)             :: Value
   END FUNCTION Integrand
END INTERFACE

Since I want to call Rule_Tn from C code I need to define a function type in C exactly matching to this interface above. Thus I tried figure out how a Fortran function returns arrays. At first I supposed the following C signature

void Integrand(double* value, const int* nf, const int* x);

matches to the interface above. But far wrong! I got a segfault. And I already tested that double is the corresponding type to REAL(KIND=STND), the STND comes out of the module Precision_Model.

Now can anyone tell me what's the right signature? I'm using the GNU Fortran and C compilers.

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