blocks|key|518018|text|我的Fortran有点(好吧，有点)生疏了，但让我试一试。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|518019|a可能收到了值，而不是指向它的指针。这会将其设置为1，这不是一个好的指针值。|offset|length|style|CODE|518020|b接收数据块中的第二个值。您的C%2B%2B结构没有向编译器提供数据的真实格式，因此它只是按照结构中给出的顺序将项分配给结构。我会得到一个指向数据块的指针，然后手工反汇编它。|518021|将a赋给数据块的地址(看起来像一个long+int指针；您的里程可能会有所不同)和b+=+a[5]。希望这能有所帮助。|518022|entityMap^0|0|0|1|P|1|0|0|1|0|1|1|H|8|15|8|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@$D|P|E|Q|F|G]|$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@$D|X|E|Y|F|G]|$D|Z|E|10|F|G]|$D|11|E|12|F|G]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|M|$]]

My Fortran is a bit (OK, QUITE a bit) rusty, but let me give it a whirl. 

<code>a</code> probably received the VALUE rather than a pointer to it. That would set it to <code>1</code>, which not a good pointer value. 

<code>b</code> received the second value in the data block. Your C++ struct gives no indication to the compiler what the real format of the data is, so it's going to just mindlessly assign the items to the structure in the order given in the struct. I'd get a pointer to the data block and disassemble it by hand. 

Assign <code>a</code> to the ADDRESS of the data block (as a <code>long int</code> pointer, looks like; your mileage may vary) and <code>b = a[5]</code>. Hope that helps.

blocks|key|1310739|text|在我看来，您的FORTRAN数据实际上布局为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1310740|struct+{
++int+a[5];
++int+b;
}|code-block|syntax|javascript|1310741|你在一台机器上，sizeof(int)+==+sizeof(int*)。|offset|length|style|CODE|1310742|entityMap^0|0|0|8|R|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Looks to me like your FORTRAN data is actually laid out as

<pre><code>struct {
 int a[5];
 int b;
}
</code></pre>

and you are on a machine where <code>sizeof(int) == sizeof(int*)</code>.

blocks|key|2156025|text|如果你想在C和Fortran之间共享全局变量，最好的方法是使用模块变量和Fortran+ISO_C_Binding。公共块是一个遗物，最好避免，除非是遗留代码的一部分。使用ISO_C_Binding将使您的代码编译器和平台独立。在gfortran手册的“混合语言编程”一章的“可互操作的全局变量”小节中有一个代码示例。这并不是特定于gfortran的，只是一些很好的文档。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2156026|继续使用ISO_C_Binding，如果您使用它提供的Fortran类型，您将确保与C类型匹配。相当于C的int的Fortran是C_INT。一个列表出现在gfortran手册的“内部模块”一章中。|offset|length|style|BOLD|2156027|entityMap^0|0|1H|3|1T|5|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

If you want to share global variables between C and Fortran, the best way is to use module variables and the Fortran ISO_C_Binding. Common blocks are a relic, best avoided unless part of legacy code. Using the ISO_C_Binding will make your code compiler and platform independent. There is a code example in the subsection "Interoperable Global Variables" of the chapter "Mixed Language Programming" of the gfortran manual. This isn't specific to gfortran, just some good documentation.

Continuing with the ISO_C_Binding, if you use the Fortran types that it provides you will be ensured of a match to the C types. The Fortran equivalent to C's int is C_INT. A list appears in the chapter "Intrinsic Modules" of the gfortran manual.

blocks|key|2156059|text|谢谢你们所有人对我的帮助。在你的建议下，我终于意识到我的问题所在。我想我的错误是在Fortran中我有：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2156060|integer+a(5),b++
common++/sample/+a,b++|code-block|syntax|javascript|2156061|使用固定大小的a(5)，然后在C%2B%2B中：|offset|length|style|CODE|2156062|extern+"C"{++
++++extern+struct{++
++++++++int+*a,b;++
}sample_;++|2156063|不带大小的*a，如指针。因此，编译器将the理解为*int指针，而不是a(5)数组|2156064|entityMap^0|0|0|7|4|0|0|5|2|P|4|Z|4|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

thak you all for your useful anwers. Following your advices I finally realized what my problem was. I think my mistake was that in Fortran I had:

<pre><code>integer a(5),b 
common /sample/ a,b 
</code></pre>

<code>a(5)</code> with a fixed size, and then in C++:

<pre><code>extern "C"{ 
 extern struct{ 
 int *a,b; 
}sample_; 
</code></pre>

<code>*a</code> without size, as a pointer. Thus, the compiler was understanding a this way, as a <code>*int</code> pointer instead of as an <code>a(5)</code> array

blocks|key|2330084|text|对于正确的C%2B%2B/Fortran代码，您应该至少使用Fortran+2003。实际上，为旧的Fortran代码(即使是Fortran+77)创建Fortran+2003/2008接口是相当简单的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2330085|这是一个关于混合现代Fortran和C%2B%2B的文章的链接：|2330086|http://solarianprogrammer.com/2012/05/11/mixed-language-programming-cpp-11-fortran-2008/|offset|length|2330087|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|2G|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@$F|R|G|S|1|T]]|A|$]]|$1|H|3|-4|5|6|7|U|8|@]|9|@]|A|$]]]|I|$J|$5|K|L|M|A|$N|E]]]]

For a proper C++/Fortran code you should use at least Fortran 2003. It is actually quite straightforward to create a Fortran 2003/2008 interface for an old Fortran code (even for Fortran 77).

Here is a link to a post about mixing modern Fortran and C++:

<a href="http://solarianprogrammer.com/2012/05/11/mixed-language-programming-cpp-11-fortran-2008/" rel="nofollow">http://solarianprogrammer.com/2012/05/11/mixed-language-programming-cpp-11-fortran-2008/</a>

I'm trying to access an array within a Fortran common block structure from C++. 
I have a mixed sample code with C++ and Fortran. 

Fortran:

<pre><code>integer a(5),b 
common /sample/ a,b 
a(1) = 1 
a(2) = 5 
a(3) = 10 
a(4) = 15 
a(5) = 20 
b = 25 
</code></pre>

Then in C++:

<pre><code>extern "C"{ 
 extern struct{ 
 int *a,b; 
}sample_; 
</code></pre>

From C++, if I try to print the value <code>sample_.b</code>: 

<pre><code>printf("sample b:%d\n",sample_.b);
</code></pre>

I get the value of a(2) : <code>sample b:5</code>

And if I try to print any other of the array-a values I just get a segementation fault...

<pre><code>printf("sample_.a[1]=%d\n",(int)sample_.a[1]); 
printf("sample_.a[0]=%d\n",(int)sample_.a[0]);
</code></pre>

What am I doing wrong?¿ Any idea ¿? 
I think, maybe I have to pass the length of the array "a" too to C++, but if so, I don't know how to do it either.

C++ and Fortran: Error accessing arrays in a Fortran common block from C++

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试从C++访问Fortran公共块结构中的数组。我有一个混合了C++和Fortran的示例代码。Fortran：integer a(5),b  common  /sample/ a,b  a(1) = 1  a(2) = 5  a(3) = 10  a(4) = 15  a(5) = 20  b = 25  然...

问C++和Fortran:从C++访问Fortran公共块中的数组时出错
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