是否可以将static_cast转换为相同类型，从而导致运行时开销? [英] Can static_cast to same type introduce runtime overhead?

问题描述

我有一个采用两种类型(T和S)的结构模板，在某些时候使用static_cast从一种类型转换为另一种类型. T和S通常是同一类型.

I have a structure template that takes two types (T and S), and at some point uses a static_cast to convert from one type to the other. It is often the case that T and S are the same type.

设置的简化示例:

template <typename T, typename S = T>
struct foo
{
  void bar(T val)
  {
    /* ... */
    some_other_function(static_cast<S>(val));
    /* ... */
  }
};

在S与T是同一类的情况下，static_cast会还是会带来额外的开销，或者它是一个空操作，将始终被忽略?

In the case that S is the same class as T, does or can the static_cast introduce extra overhead, or is it a null operation which will always be ignored?

如果确实增加了开销，是否有一个简单的模板元编程技巧仅在需要时才执行static_cast，还是我需要创建部分专业化知识来应对T == S情况?如果可能的话，我宁愿避免整个foo模板的部分专业化.

If it does introduce overhead, is there a simple template metaprogramming trick to perform the static_cast only if needed, or will I need to create a partial specialization to cope with the T == S case? I'd rather avoid the partial specialization of the entire foo template if possible.

推荐答案

是的，可以.

这里是一个例子:

struct A {
  A( A const& ) {
    std::cout << "expensive copy\n";
  }
};

template<typename T>
void noop( T const& ) {}
template <typename T, typename S = T>
void bar(T val)
{
  noop(static_cast<S>(val));
}
template <typename T>
void bar2(T val)
{
  noop(val);
}
int main() {
  std::cout << "start\n";
  A a;
  std::cout << "bar2\n";
  bar2(a); // one expensive copy
  std::cout << "bar\n";
  bar(a); // two expensive copies
  std::cout << "done";
}

基本上，static_cast可以导致调用副本构造函数.

basically, a static_cast can induce a copy constructor to be called.

对于某些类型(例如int)，复制构造函数基本上是免费的，编译器可以消除它.

For some types (like int), a copy constructor is basically free, and the compiler can eliminate it.

对于其他类型，不能.在这种情况下，复制省略也不合法:如果您的复制构造函数有副作用，或者编译器无法证明它没有副作用(通常，如果复制构造函数很重要)，它将被调用.

For other types, it cannot. In this context, copy elision isn't legal either: if your copy constructor has side effects or the compiler cannot prove that it has no side effects (common if the copy constructor is non-trivial), it will be called.

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