Question

我正在尝试遍历Player对象的向量，然后根据Player指针返回下一个对象。我已经尝试了各种方法来使lambda中的运算符匹配，但无法使其工作并在堆栈溢出时读取类似的此类问题。我很抱歉没有SSCCE，我只是不知道我目前的技能水平。

c_player的来源：
定义为：

Player* current_player;

设置为：

void Game::set_first_player_turn(){
    current_player = &game_players.back();
}

错误的代码：

using namespace std;
vector<Player>game_players;

Player* Game::find_player(Player* c_player){
    vector<Player>::iterator iter;
    iter = find_if(game_players.begin(), game_players.end(),
                   [&](Player* p ) -> bool{ return p == c_player; }); //This line
                                        // causes the exception. Sets iterator to
                                        // position of the current player
    if (iter != game_players.end()){
        advance(iter, 1);
    }
    else{
        iter == game_players.begin();
    }
    current_player = &(*iter);  

    return current_player;
}

编译错误：

error C2664: 'bool Game::set_game_flow::<lambda_360ac8a355100cfac1adc9f8eba8d8b9>
::operator ()(Player *) const' : cannot convert parameter 1 from 'Player' 
to 'Player *'

感谢您的帮助。

Answer 1

find_if中的谓词必须对存储在容器中的类型进行const引用。您在向量中存储Player类型

vector<Player> game_players;

因此

[&](Player* p )

不是正确的谓词。这应该是

[c_player]( const Player& p ) { return p == *c_player; } 
                                    // ^^^^^^^^^^^^^^
                                    // to compare by value
// note: to compare by value 
// bool operator==( const Player&, const Player&)
// must be defined

[c_player]( const Player& p ) { return &p == c_player; } 
                                    // ^^^^^^^^^^^^^^
                                    // to compare by address
// note: address of c_player
// has to be valid (not invalidated by vector through resizing, erasing, etc)

所以你可以写：

vector<Player>::iterator iter;
iter = find_if( game_players.begin(), game_players.end(),
                       [c_player]( const Player& p ) { return p == *c_player;});

基于对象指针在向量内获取对象

1 个答案: