Visual Studio 2015“非标准语法;使用'&'创建指向成员的指针“

时间:2016-06-14 15:00:54

标签: c++ visual-studio-2015

我正在学习C ++并尝试制作Tic Tac Toe的小游戏。但我继续得到C3867,非标准语法;使用'&'创建一个记住的指针。

这是我的TicTacToe.h:

#pragma once
#include <iostream>

using namespace std;

class TicTacToe
{
public:
    TicTacToe();
    string getName1();
    string getName2();
    void printBoard();
    void clearBoard();
    void setName1(string player1Name);
    void setName2(string player2Name);
    void setSign1(string player1Sign);
    void setSign2(string player2Sign, string player1Sign);
    void playGame(string player1Name, string player2Name,string    player1Sign,string player2Sign);                  
    void player1Move(string coordX);
    void player1Turn();
    void player2Turn();
private:
    char Board[3][3];
    string _player1Name;
    string _player2Name;
    string _player1Sign;
    string _player2Sign;
    string _coordX;
    string _coordY;
};

这是我的TicTacToe.cpp:

#include "TicTacToe.h"
#include <iostream>
#include <string>

TicTacToe::TicTacToe() {}

void TicTacToe::playGame(string player1Name, string player2Name,
                         string player1Sign, string player2Sign) {
  TicTacToe Board;
  Board.setName1(player1Name);
  Board.setSign1(player1Sign);
  Board.setName2(player2Name);
  Board.setSign2(player1Sign, player2Sign);
  Board.clearBoard();
  Board.printBoard();
}

void TicTacToe::printBoard() {
  cout << " |1|2|3|\n";
  for (int i = 0; i < 3; i++) {
    cout << "--------\n";
    cout << i + 1 << "|" << Board[i][0] << "|" << Board[i][1] << "|"
         << Board[i][2] << "|" << endl;
  }
}

void TicTacToe::clearBoard() {
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      Board[i][j] = ' ';
    }
  }
}

void TicTacToe::setName1(string player1Name) {
  cout << "Enter your name, player 1: \n";
  cin >> player1Name;
  _player1Name = player1Name;
}

void TicTacToe::setName2(string player2Name) {
  cout << "Enter your name, player 2: \n";
  cin >> player2Name;
  _player2Name = player2Name;
}

string TicTacToe::getName1() { return _player1Name; }

string TicTacToe::getName2() { return _player2Name; }

void TicTacToe::setSign1(string player1Sign) {
  cout << "What will you sign be?(X/O)\n";
  cin >> player1Sign;
  if (player1Sign != "X" && player1Sign != "O" && player1Sign != "x" &&
      player1Sign != "o") {
    cout << "Invalid input, try again.\n";
    cin >> player1Sign;
  }
  _player1Sign = player1Sign;
}

void TicTacToe::setSign2(string player2Sign, string player1Sign) {
  cout << "What will you sign be?(X/O)\n";
  cin >> player2Sign;

  if (player2Sign != "X" && player2Sign != "O" && player2Sign != "x" &&
          player2Sign != "o" ||
      player2Sign == player1Sign) {
    cout << "Invalid input, try again.\n";
    cin >> player2Sign;
  }
  _player2Sign = player2Sign;
}

void TicTacToe::player1Move(string coordX) // ERROR
{
  cout << "Enter X: " << endl;
  cin >> coordX;
  _coordX = coordX;
}

void TicTacToe::player1Turn() {
  cout << "Player 1 turn !\n";
  TicTacToe Board;
  Board.player1Move;
}

void TicTacToe::player2Turn() {
  cout << "Player 2 turn !\n";
  TicTacToe Board;
  Board.player1Move;
}

我在其他问题上尝试过关于此错误的所有内容,但它们无效。你如何解决这个错误?

3 个答案:

答案 0 :(得分:24)

answer by drorco已经提供了对您的问题的修复。我将尝试解释错误消息。

如果您有非成员函数,则可以在不使用函数调用语法的情况下在表达式中使用函数名称。

void foo()
{
}

foo; // Evaluates to a function pointer.

但是,当你有一个成员函数时,在没有函数调用语法的表达式中使用成员函数名是无效的。

struct Bar
{
   void baz() {}
};

Bar::baz;  // Not valid.

要获取指向成员函数的指针,您需要使用&运算符。

&Bar::baz;   // Valid

这解释了Visual Studio的错误消息:

"non-standard syntax; use '&' to create a pointer to member"

答案 1 :(得分:9)

问题在于包含以下内容的行(在代码中出现两次):

Board.player1Move;

播放器一次移动是一个接收std :: string参数作为输入的函数。为了调用它,你需要创建一个std :: string对象并将其作为函数的参数传递。如果希望将字符串作为输入提供,则可以使用以下语法:

std::string move;
cin >> move;
Board.player1Move(move);

另外,请注意,player2Turn应该调用Board.player2Move而不是Board.player1Move。

答案 2 :(得分:2)

我快速将std::exception添加到try/catch代码块的应用中,我就是这样做的,如下所示:

try{
//code block
}catch(std::exception e){printf("ERROR: %s", e.what()); return -1;}

我只需要将例外的参数更改为以&开头,如下所示:

try{
//code block
}catch(std::exception &e){printf("ERROR: %s", e.what()); return -1;}