c++ - CUDA: "missing return statement at end of non-void function"in constexpr if 函数

Question

当我编译以下测试代码时，我收到此警告:

test.cu(49): warning: missing return statement at end of non-void function "AllocateSize<T,D>(size_t) noexcept [with T=int, D=Device::GPU]"
          detected during instantiation of "Pointer<T, D> AllocateSize<T,D>(size_t) noexcept [with T=int, D=Device::GPU]" 
(61): here

我应该担心吗，这是预期的吗？我该怎么做才能让它消失？这看起来很奇怪，因为 cuda 确实支持 C++17。提前致谢!
编译: nvcc -std=c++17 test.cu -o test测试代码( test.cu ):

enum class Device { CPU, GPU }; // Device

template <typename T, Device D>
class Pointer {
private:
    T* m_raw = nullptr;
    
public:
    __host__ __device__ inline Pointer(T* const p)              noexcept { this->SetPointer(p); }

    __host__ __device__ inline void SetPointer(const Pointer<T, D>& o) noexcept { this->m_raw = o.m_raw; }

    template <typename U>
    __host__ __device__ inline Pointer<U, D> AsPointerTo() const noexcept {
        return Pointer<U, D>(reinterpret_cast<U*>(this->m_raw));
    }

    __host__ __device__ inline operator T*& () noexcept { return this->m_raw; }
}; // Pointer<T, D>

template <typename T>
using CPU_Ptr = Pointer<T, Device::CPU>;

template <typename T>
using GPU_Ptr = Pointer<T, Device::GPU>;

template <typename T, Device D>
__host__ inline Pointer<T, D> AllocateSize(const size_t size) noexcept {
    if constexpr (D == Device::CPU) {
        return CPU_Ptr<T>(reinterpret_cast<T*>(std::malloc(size)));
    } else {
        T* p;
        cudaMalloc(reinterpret_cast<void**>(&p), size);
        return GPU_Ptr<T>(p);
    }
}

template <typename T, Device D>
__host__ inline void Free(const Pointer<T, D>& p) noexcept {
    if constexpr (D == Device::CPU) {
        std::free(p);
    } else {
        cudaFree(p.template AsPointerTo<void>());
    }
}

int main() { Free(AllocateSize<int, Device::GPU>(1024)); }

CUDA 版本 11.1

基于 Ubuntu 的 Linux 发行版

Answer 1

我们的编译器团队研究了这个问题。该警告是虚假警告。我们希望在 future 的 CUDA 版本中解决这个问题。我将无法回答有关何时可能出现的问题。代码正在正确生成。
如果您希望消除警告，一种可能的方法是在相关函数的末尾添加额外的 return 语句。在这种情况下，这应该不起作用，因为它无法访问:

template <typename T, Device D>
__host__ inline Pointer<T, D> AllocateSize(const size_t size) noexcept {
    if constexpr (D == Device::CPU) {
        return CPU_Ptr<T>(reinterpret_cast<T*>(std::malloc(size)));
    } else {
        T* p;
        cudaMalloc(reinterpret_cast<void**>(&p), size);
        return GPU_Ptr<T>(p);
    }
    return Pointer<T, D>(NULL);
}

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