我正在学习 C++,并尝试制作一个井字游戏。但我总是得到 C3867,非标准语法;使用'&'创建一个指针来记忆。
这是我的 TicTacToe.h .NET:
#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;
};
这里是我的 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;
}
我已经尝试了有关此错误的所有其他问题,但它们都不起作用。您是如何修复这个错误的?
drorco 的回答](https://stackoverflow.com/a/37815886/434551) 中已经提供了解决问题的方法。下面我试着解释一下错误信息。
当您有一个非成员函数时,您可以在表达式中使用函数名,而无需使用函数调用语法。
void foo()
{
}
foo; // Evaluates to a function pointer.
但是,当使用成员函数时,在表达式中使用成员函数名而不使用函数调用语法是无效的。
struct Bar
{
void baz() {}
};
Bar::baz; // Not valid.
要获取指向成员函数的指针,需要使用 &
操作符。
&Bar::baz; // Valid
这就是 Visual Studio 错误信息的原因:
"non-standard syntax; use '&' to create a pointer to member"
问题出在包含以下内容的行中(在您的代码中出现了两次):
Board.player1Move;
Player one move 是一个接收 std::string 参数作为输入的函数。为了调用该函数,您需要创建一个 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,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。
当我快速将 std::exception
添加到应用程序的 try/catch
代码块时,就会出现这种情况,就像这样:
try{
//code block
}catch(std::exception e){printf("ERROR: %s", e.what()); return -1;}
我只需将异常名称的参数改为以&
开头,就像这样:
try{
//code block
}catch(std::exception &e){printf("ERROR: %s", e.what()); return -1;}