接受模板作为函数参数

Question

Could someone explain the difference between the two function calls here?

When can you pass a templated variable to a function?

In both cases I should end up with an templated array inside the function, but only one compiles.

template<int DIM>
struct MyStruct
{
    array<int, DIM> structArr;
};

template<int DIM> void testA( MyStruct  <DIM>& myStruct)    {    }

template<int DIM> void testB( array<int, DIM>& arrA)        {    }

int main()
{
    MyStruct<3>     myStruct;
    array<int, 3>   arr;

    testA(myStruct);
    testB(arr);         //compile error

    return 0;
}

EDIT: Error messages look like this:

error: no matching function for call to ‘testB(std::array&)’
 testB(arr);         //compile error
          ^
note: candidate: template void testB(std::array&)
 template<int DIM> void testB( array<int, DIM>& arrA)        {    }
                        ^~~~~
note:   template argument deduction/substitution failed:
note:   mismatched types ‘int’ and ‘long unsigned int’

Answer 1

The template parameter for the size of a std::array is of type std::size_t . However, here you require supplying an int for DIM in the function definition. This is probably an issue for the template deduction rules.