call cudaCreateChannelDesk from template class how to tell compiler that cudaCreateChannelDesk is te

Accelerated Computing CUDA CUDA Programming and Performance
1

GCC will not compile following code snippet (which in fact is correct behavior of GCC as it is conform to the standard c++)

template<class T>

void CUDAMemory1D2DTextureAllocator<T>::allocateMemoryOnDevice()

{

    m_pChannelDesc = cudaCreateChannelDesc<T>();    // this gives compile error in gcc (msvc compiles fine)

    cudaError_t result = cudaMallocArray( &m_pDeviceMemory, &m_pChannelDesc, m_pWidth, m_pHeight);

    ...

}

One needs to tell the compiler that cudaCreateChannelDesc is a template method. Otherwise it will try to

parse < as a smaller than operator…

The following snippet shows that in an easy example:

template< typename G >

struct Test

{

    template< typename T > T f() const;

};

template< typename G, typename T >

 void g()

 {

     Test< G > t;

t.f< T >();           // ERROR: gcc won't compile that

     t.template f< T >();  // OK: now gcc knows that f is a template method an treads the following angle brackets not as operators but as template brackets...

 }

So far so good. Now my question is, how to do that in the above case, where the method i call is cudaCreateChannelDesc which does not belong to any class or namespace???

Any advice or suggestion how to solve this situation are very welcome.

Thanks

Tobi