c++ - 尝试使用模板函数交换两个字符串

Question

#include<iostream>
#include<string>

template <typename T>
void swap(T a , T b)
{
  T temp = a;
  a = b;
  b = temp;
}

template <typename T1>
void swap1(T1 a , T1 b)
{
  T1 temp = a;
  a = b;
  b = temp;
}

int main()
{
  int a = 10 , b = 20;
  std::string first = "hi" , last = "Bye";

  swap(a,b);
  swap(first, last);   

  std::cout<<"a = "<<a<<" b = "<<b<<std::endl;
  std::cout<<"first = "<<first<<" last = "<<last<<std::endl;    

  int c = 50 , d = 100;
  std::string name = "abc" , surname = "def";

  swap1(c,d);
  swap1(name,surname);

  std::cout<<"c = "<<c<<" d = "<<d<<std::endl;
  std::cout<<"name = "<<name<<" surname = "<<surname<<std::endl;    

  swap(c,d);
  swap(name,surname);

  std::cout<<"c = "<<c<<" d = "<<d<<std::endl;
  std::cout<<"name = "<<name<<" surname = "<<surname<<std::endl;    

  return 0;
}

**Output**
a = 10 b = 20
first = Bye last = hi
c = 50 d = 100
name = abc surname = def
c = 50 d = 100
name = def surname = abc

swap()和 swap1()基本上都具有相同的功能定义，那么为什么只有 swap()实际交换字符串，而 swap1()没有？

还可以告诉我默认情况下如何将STL字符串作为参数传递，即它们是按值还是按引用传递？

Answer 1

我明白了为什么人们现在不喜欢ADL ...

您看到的是Argument Dependent Lookup的效果。如果要在swap实现中添加打印件，您会注意到，它是而不是std::string的，而仅是int的。

与您的版本相比，首选std::swap，因为存在std::basic_string类型的explicit specialization。如果不存在，则通话可能会模棱两可。
对于int，在查找过程中不考虑 namespace std，因此您的版本是唯一可接受的版本。

Also can you tell me that how are stl strings passed as arguements by default i.e are they passed by value or by reference?

C++中的所有内容均按值传递，除非您将其明确标记为按引用传递。