为什么C ++参数范围影响命名空间中的函数查找？ [英] Why does C++ parameter scope affect function lookup within a namespace?

问题描述

这似乎有点向后，但它的工作原理：

This seems a little backwards to me but it works:

#include <iostream>

namespace nTest
{
  struct cTest {};

  void fTest(cTest& x)
  {
    std::cout << "nTest::fTest(cTest&) called" << std::endl;
  }
}

int main(void)
{
  nTest::cTest x;
  fTest(x); //Weird! fTest is resolved since its parameter belongs to nTest.
  return 0;
}

通常，您需要nTest ::才能访问fTest，属于nTest的参数似乎将nTest添加到搜索fTest的可能范围的列表中。对我来说，参数范围影响函数查找似乎很奇怪。

Normally, you would need nTest:: in order to access fTest, but its parameter which belongs to nTest appears to add nTest to the list of possible scopes in which to search for fTest. It seems odd to me that the parameter scope influences the function lookup.

这在GCC中编译得很好，但我想知道这个用法是可移植的吗？

This compiles fine in GCC, but I'm wondering is this usage portable? What is the official definition of this scoping mechanism?

推荐答案

这是ADL（Argument Dependent Lookup）或Koenig Lookup的特征）。该特性的目的是在许多情况下，相同的命名空间将包含可应用于这些类型的类型和函数，所有这些类型都符合 界面 。如果ADL不在位，您必须使用使用 c>声明将标识符带入范围，否则您必须限定调用。

That is ADL (Argument Dependent Lookup) or Koenig Lookup (for the designer of the feature). The purpose of the feature is that in many cases the same namespace will contain types and functions that can be applied to those types, all of which conform the interface. If ADL was not in place, you would have to bring the identifiers into scope with using declarations or you would have to qualify the calls.

这成为一个噩梦，因为语言允许运算符重载。请考虑以下示例：

This becomes a nightmare since the language allows for operator overloads. Consider the following example:

namespace n {
   struct test {};
   test operator+( test, test const & ); // implemented
};
int main() {
   n::test a,b;
   n::test c = a + b;  //without ADL: c = n::operator+( a, b )
}

虽然它可能看起来像一个尴尬的情况，考虑 n 可能是 std 命名空间，测试可以是 ostream 和 operator 可以是 < ：

While it might seem like an awkward situation, consider that n might be the std namespace, test might be ostream, and operator+ could be operator<<:

int main( int argc, char** ) {
   std::cout << "Hi there, there are " << argc << " arguments" << std::endl;
}

没有ADL时，调用 operator< 必须是显式的，此外，你必须知道哪些是实现为自由函数对方法。你知道 std :: cout<< Hi正在调用一个自由函数， std :: cout< 5 正在调用成员函数？没有多少人意识到，认真，几乎没有人关心。 ADL隐藏了你。

Without ADL, the calls to operator<< would have to be explicit, and moreover you would have to know which of them is implemented as a free function versus a method. Did you know that std::cout << "Hi" is calling a free function and std::cout << 5 is calling a member function? Not many people realize it, and seriously, almost no one cares. ADL hides that from you.

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