在C ++中对Fortran函数的未定义引用 [英] Undefined reference to Fortran function in C++

问题描述

我似乎无法找出为什么这不工作。

I can't seem to figure out why this isn't working.

/* main.cpp */
#include <stdio.h>
extern "C"
{
    int __stdcall inhalf(int *);
}
int main()
{
    int toHalf = 2;
    int halved = inhalf(&toHalf);
    printf("Half of 2 is %d", halved);
    return 0;
}

好，看起来不错。

$ g++ -c main.cpp

没有错误。

! functions.f90
function inhalf(i) result(j)
    integer, intent(in) :: i
    integer             :: j
    j = i/2
end function inhalf

我相信这是对的。

$ gfortran -c functions.f90

太好了...

$ gcc -o testo main.o functions.o
main.o:main.cpp:(.text+0x24): undefined reference to `inhalf@4'
collect2.exe: error: ld returned 1 exit status

我一直在寻找这个超过一个小时，但我找不到任何工作的情况下。我应该如何解决这个问题？

I've been looking this up for over an hour, but I couldn't find anything that worked for this case. How should I solve this?

推荐答案

对于完整的C兼容性你可以使用 bind 现代Fortran的功能：

For full C compatibility you can use the bind feature of modern Fortran:

! functions.f90
function inhalf(i) result(j) bind(C,name='inhalf')
    integer, intent(in) :: i
    integer             :: j
    j = i/2
end function inhalf

这样可以给函数，您可以在C（和其他人）中使用，而不依赖于编译器自己使用的命名方案。

This allows you to give a name to the function that you can use in C (and others) without relying on the naming scheme your compiler uses on its own.

__ stdcall 仅适用于Win32（以及链接的默认行为，请参阅在这里）。您可以安全地删除它。 [实际上，在Linux中编译代码是必需的。 ]

The __stdcall is Win32 only (and the default behavior for linking, see here). You can safely remove it. [Actually it is required for compiling your code in Linux. ]

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