从 C++ function 转换返回值

Question

I have following C++ code snippet

inline std::vector<std::unique_ptr<xir::Tensor>> cloneTensorBuffer(
    const std::vector<const xir::Tensor*>& tensors) {
  auto ret = std::vector<std::unique_ptr<xir::Tensor>>{};
  auto type = xir::DataType::XINT;
  ret.reserve(tensors.size());
  for (const auto& tensor : tensors) {
    ret.push_back(std::unique_ptr<xir::Tensor>(xir::Tensor::create(
        tensor->get_name(), tensor->get_shape(), xir::DataType{type, 8u})));
  }
  return ret;
}

I am not clear about the expression:

std::unique_ptr<xir::Tensor>(xir::Tensor::create(
            tensor->get_name(), tensor->get_shape(), xir::DataType{type, 8u}))

Is the expression casting the value returned by xir::Tensor::create() to std::unique_ptr<xir::Tensor? I am confused since the C++ casting syntax is (type)expression Can someone explain please.

regards, -sunil puranik

Answer 1

This code takes a vector of raw pointers to Tensor objects, and returns a new vector of unique pointers to a new set of Tensor objects. Doesn't look like any explicit casting is taking place.

std::unique_ptr<xir::Tensor>( // create unique pointer to new Tensor object
    xir::Tensor::create(
        tensor->get_name(),  // created with name and shape of the
        tensor->get_shape(), // tensor pointed at in the original vector
        xir::DataType{type, 8u} // and with unsigned 8-bit XINT datatype
    )
)

Regarding casting - please don't consider eg (int)foo to be C++-style casting - that's legacy C-style casting. Amongst other things it's hard to find by searching a codebase!

C++ has:

• static_cast
• dynamic_cast
• reinterpret_cast
• const_cast
• implicit conversion

There's more detail on this at cplusplus.com .