CODEWITHSUNDEEP
c++templates
Ran Wang
Ran Wang

Reputation: 303

Error in using templated function with template parameter to be templated class

here is a minimal example

#include <iostream>

template<typename T>
struct MyClass{
    T value;
};

template<typename T, template<typename> class Class>
void foo(Class<T>& myClass){
    myClass.value=0;
};

int main(){
    MyClass<double> myclass;
    foo<double, MyClass<double>>(myclass);
    return 0;

}

This code will not compile and gives the error

 error: no instance of function template "foo" matches the argument list
            argument types are: (MyClass<double>)
      foo<double, MyClass<double>>(myclass);

The reason why I want to write a function is that I want to write a function that transfer data between CPU and GPU. The function looks like

template<typename Scalar, template<typename > class Host,
        template<typename> class Device>
void copyFromHostAsync(const Host<Scalar> &host, Device<Scalar> &device, cudaStream_t stream) {
    assert(host.rows() == device.rows());
    assert(host.cols() == device.cols());

    cudaMemcpyAsync(device.getPtr(), host.getPtr(), sizeof(Scalar) * host.rows() * host.cols(),
                    cudaMemcpyHostToDevice, stream);

}

I want to use the templated class as parameter so that the underlying scalar type is the same.

Any help is welcomed.

Upvotes: 1

Views: 44

Answers (1)

Rakete1111
Rakete1111

Reputation: 49028

foo is a template function, which takes as template parameter a type T, and a template type with 1 argument type MyClass.

If you write:

foo<double, MyClass<double>>(myclass);

The second template parameter is not a template with 1 parameter type. It is just a simple type. And because it is a type, not a template type, your code doesn't compile.

Using just MyClass will compile, as MyClass is a template type which takes 1 template parameter:

foo<double, MyClass>(myclass);

Also, just let the compiler do the job for you and let it deduce the types:

foo(myclass);

Upvotes: 2

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