c++ - 单行在 map 中初始化C++对象

Question

初始化类并直接分配给unordered_map的正确方法是什么？

#include <string>

#include <unordered_map>
int main() {
  std::unordered_map<std::string, Foo> s;
  // Foo foo{1};
  s["test"] = Foo(1); // this is bad
  return 0;
}

oo

class Foo {
  public:
    Foo(int x)
      : x_(x) {}

  private:
    int x_;
};

现在我明白了

main.cpp:19:4: note: in instantiation of member function 'std::unordered_map<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, Foo, std::hash<std::__cxx11::string>, std::equal_to<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::allocator<std::pair<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, Foo> > >::operator[]' requested here
  s["test"] = Foo(1);
   ^
Foo.h:3:5: note: candidate constructor not viable: requires single argument 'x', but no arguments were provided
    Foo(int x)
    ^
Foo.h:1:7: note: candidate constructor (the implicit copy constructor) not viable: requires 1 argument, but 0 were provided
class Foo {
      ^
Foo.h:1:7: note: candidate constructor (the implicit move constructor) not viable: requires 1 argument, but 0 were provided

问题在于 s["test"] = Foo(1)。

Answer 1

当您想使用operator[]时，映射中的元素必须是默认可构造的，因为当映射中尚不存在给定键的值时，它将默认构造该值。使您的Foo默认为可构造的，或改用 insert 或 emplace 。

默认可构造的Foo:

class Foo {
  public:
    Foo(int x = 0)   // <- can be called without parameters
      : x_(x) {}

  private:
    int x_;
};