将字符串推送到私有成员向量

Question

我有一个std::vector<std::string> tileSet_作为class的成员变量。在上述 class 的构造函数中，我调用函数

void MapWindow::shuffleTiles() const
{
  std::vector<std::pair<std::string, unsigned>> tileAmounts = {
    {"a", 2}, {"b", 4}, {"c", 1}, {"d", 4},
    {"e", 5}, {"f", 2}, {"g", 1}, {"h", 3},
    {"i", 2}, {"j", 3}, {"k", 3}, {"l", 3},
    {"m", 2}, {"n", 3}, {"o", 2}, {"p", 3},
    {"q", 1}, {"r", 3}, {"s", 2}, {"t", 1},
    {"u", 8}, {"v", 9}, {"w", 4}, {"x", 1}
  };

  int tileN = 0;
  for (std::vector<std::pair<std::string, int>>::size_type i = 0;
    i < tileAmounts.size();i++) {
    tileN += tileAmounts.at(i).second;
  }

  // only used once to initialise (seed) engine
  std::random_device rd;
  // random-number engine used (Mersenne-Twister in this case)
  std::mt19937 rng(rd());
  // guaranteed unbiased
  std::uniform_int_distribution<int> uni(0,tileAmounts.size() - 1);

  for (int i = 0; i < tileN; i++) {
    auto random_integer = uni(rng);
    //qDebug() << i << ":" << random_integer;
    std::cout << i
              << ":"
              << tileAmounts.at(random_integer).first
              << ":"
              << tileAmounts.at(random_integer).second
              << std::endl
              << std::endl;

    while (tileAmounts.at(random_integer).second < 1) {
      std::cout << "Reshuffling "
               << tileAmounts.at(random_integer).first
               << ":"
               << tileAmounts.at(random_integer).second
               << std::endl;
      random_integer = uni(rng);
    }

    std::string s = tileAmounts[random_integer].first;
    std::cout << s
              << std::endl;

    tileSet_.push_back(s);
    tileAmounts.at(random_integer).second -= 1;
  }

  for (std::vector<std::string>::size_type i = 0;
    i < tileSet_.size();i++) {
    std::cout << tileSet_.at(i)
              << std::endl;
  }
}

它在最后尝试将向量 tileAmounts 中包含的一对随机选取的字符串推回到 memeber 向量中。出于某种原因，这引发了错误

no matching member function call to 'push_back'

为什么会这样。我正在将矢量从 std::pair 推回，但肯定不会导致任何类型不匹配？ std::vector中包含的std::pair中的std::string仍然是std::string，不是吗？

Answer 1

我找到了问题的根源：vectortileSet_ 以前不是成员变量，但我出于需要才这样做的。但是，我最初将函数 shuffleTiles 定义为 const 函数，这当然使它无法更改 class 的状态。从定义中删除 const 解决了这个问题。

C++ 中的错误消息肯定是神秘的。

Answer 2

我花了一些时间才完成所有的构建（有一些猜测）。

问题是该方法被标记为const。因此，您不允许在成员上调用任何可变函数（即您不能在任何成员变量上调用 none const 成员函数或 none const 成员函数）。

 void MapWindow::shuffleTiles() const
                                ^^^^^
 ...
     tileSet_.push_back(s);  // Attempt to mutate the state of the object.

这在编译器生成的错误消息中有更详细的解释：

t2.cpp:61:14: error: no matching member function for call to 'push_back'
    tileSet_.push_back(s);
    ~~~~~~~~~^~~~~~~~~
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/vector:712:36: note: candidate function not viable: 'this' argument has type
      'const std::vector<std::string>' (aka 'const vector<basic_string<char, char_traits<char>, allocator<char> > >'), but method is not marked const
    _LIBCPP_INLINE_VISIBILITY void push_back(const_reference __x);
                                   ^
/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/vector:715:36: note: candidate function not viable: 'this' argument has type
      'const std::vector<std::string>' (aka 'const vector<basic_string<char, char_traits<char>, allocator<char> > >'), but method is not marked const
    _LIBCPP_INLINE_VISIBILITY void push_back(value_type&& __x);

如果您转到行尾，它会显示：“但方法未标记为常量”

解密行：

note: candidate function not viable: 'this' argument has type
'const std::vector', but method is not marked const
void push_back(const_reference __x);

note: candidate function not viable: 'this' argument has type
'const std::vector', but method is not marked const
void push_back(value_type&& __x);

将字符串推送到私有成员向量

Pushing a string to a private member vector

c++

compiler-errors

vector

push-back

c++14