Convert .Net ref (%) to native (&)

Question

How can I convert a C++/CLI int %tmp to native C++ int &tmp?

void test(int %tmp)
{
    // here I need int &tmp2 for another pure C++ function call
}

score 3 · Answer 1 · answered Jun 07 '11 at 12:53

Neither of the existing answers properly handle in/out parameters, let alone any advanced use cases.

This should work for all cases where other_func does not keep the reference after it returns:

void test(int %tmp)
{
    pin_ptr<int> pinned_tmp = &tmp;
    other_func(*pinned_tmp);
}

score 1 · Answer 2 · answered Jun 06 '11 at 09:30

Just tried this, works fine:

  //in the C++ dll
void testFunc( int& n )
{
  n = 5;
}

  //in the CLI app
[DllImport( "my.dll", EntryPoint = "?exported_name_here",
   CallingConvention = CallingConvention::StdCall )]
void TestFunc( int& );

void test( int% tmp ) 
{
  int n;
  TestFunc( n );
  tmp = n;
}

Doc Brown · Accepted Answer · 2011-06-06T09:41:30.670

0

void your_function(int *);
void your_function2(int &);

void test(int %tmp)
{
    int tmp2;
    your_function(&tmp2);
    your_function2(tmp2);
    tmp=tmp2;
}

edited Jun 06 '11 at 09:41

answered Jun 06 '11 at 09:15

Doc Brown

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this will almost certainly give an error: compiler cannot convert from int* to int& – stijn Jun 06 '11 at 09:24
@stijn: That depends on the signature of "your_func". See my changed example. – Doc Brown Jun 06 '11 at 09:39

Convert .Net ref (%) to native (&)

3 Answers3