Reputation: 2224
I am in the need of using some functions in a C made program. To test I defined the following :
This is my .h file :
namespace amt
{
class AMT_EXPORT FaceRecognition
{
public:
std::string amt_test_string(std::string in);
};
};
This is my .cpp file :
#include <memory.h>
#include <string>
#include <iostream>
#include <fstream>
#include "api_shared.h"
#include <sys/stat.h>
using namespace std;
std::string amt::FaceRecognition::amt_test_string (std::string in)
{
std::string s="in: "+in;
std::cout<<s<<std::endl;
return s;
}
I am trying to invoke the method like this :
const string str = "C:\\minimal.dll";
[DllImport(str)]
public static extern string amt_test_string(string input);
static void Main(string[] args)
{
string myinput = "12";
string myoutput = "";
myoutput = amt_test_string(myinput);
Console.WriteLine(myoutput);
Console.Read();
}
But im getting an error saying that it cannot find the entry point named amt_test_string..why so? I am a newbie in C btw
Upvotes: 0
Views: 110
Reputation: 400274
That's not a C DLL, that's a C++ DLL. C and C++ are not the same language. In particular, C++ has name mangling, so the function name which gets exported to the DLL is decorated.
I'd strongly recommend that you avoid having C++ exports in your DLL for that reason. If you use only C exports, the symbol name will be predictable (i.e. will not depend on the specific details of how your C++ compiler decorates names), and you won't have to worry about runtime differences, like how your C++ standard library implements std::string
.
I'd recommend your DLL export look like this:
extern "C" // This says that any functions within the block have C linkage
{
// Input is 'in', output gets stored in the 'out' buffer, which must be 'outSize'
// bytes long
void DLLEXPORT amt_FaceRecogniztion_amt_test_string(const char *in, char *out, size_t outSize)
{
...
}
}
This interface does not rely on any particular library's std::string
implementation, and C# knows how to martial char*
parameters as C strings. However, memory management is more complicated, as you need to figure out an upper bound on how big the output is going to be and pass in an appropriately sized bufer.
Upvotes: 3