因此,我试图制作一个井字游戏,并找到了一个使用cin.get()和cin.ignore()暂停程序的示例代码。我发现关于这两个问题的解释是他们做什么。
到目前为止,我对这两个函数的了解是,cin.get()可以用来抓取变量的第一个字母,cin.ignore()至少会忽略变量的第一个字符。
以下是我找到的样本:
// istream::ignore example
#include <iostream> // std::cin, std::cout
int main () {
char first, last;
std::cout << "Please, enter your first name followed by your surname: ";
first = std::cin.get(); // get one character
std::cin.ignore(256,' '); // ignore until space
last = std::cin.get(); // get one character
std::cout << "Your initials are " << first << last << '\n';
return 0;
}
这是我未完成的井字游戏程序,在第66行和第67行使用.ignore和.get(有//****......),当程序请求移动时输出“无效输入”时,它用来暂停程序。
#include <iostream>
#include <cstdlib>
#include <string>
using namespace std;
string k;
int scoreP1, scoreP2;
char cell[10] = { ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' '};
void board(/*int scoreP1, int scoreP2*/);
void piecePlacer(int& pMover, char& piece);
int main()
{
int player, i = 1, pMove, choice, j;
char piece;
do{
cout << "Tic-Tac-Toe\n\n"
<< "[1] Player vs Player\n"
<< "[2] Player vs Computer\n"
<< "[3] Exit";
cin >> choice;
while (choice == 1){
do{
system("cls"); // fix (change)
board(/*0,0*/);
player = (i % 2) ? 1 : 2; // when i % 2 == 1 (true), player = 1; when i % 2 == 0 (false), player = 2
piece = (player == 1) ? 'X' : 'O';
cout << "\n\t Player " << player << ", it's your turn ";
cin >> pMove;
piecePlacer(pMove, piece);
i++;
} while (pMove == 0); // piecePlacer initializes pMove to 0 when user enters invalid number
}
} while (choice != 3);
return 0;
}
void piecePlacer(int& pMove, char& piece){
if (pMove == 1 && cell[1] == ' ')
cell[1] = piece;
else if (pMove == 2 && cell[2] == ' ')
cell[2] = piece;
else if (pMove == 3 && cell[3] == ' ')
cell[3] = piece;
else if (pMove == 4 && cell[4] == ' ')
cell[4] = piece;
else if (pMove == 5 && cell[5] == ' ')
cell[5] = piece;
else if (pMove == 6 && cell[6] == ' ')
cell[6] = piece;
else if (pMove == 7 && cell[7] == ' ')
cell[7] = piece;
else if (pMove == 8 && cell[8] == ' ')
cell[8] = piece;
else if (pMove == 9 && cell[9] == ' ')
cell[9] = piece;
else{
cout << "\n\t\t Invalid Move.";
pMove = 0;
cin.ignore(); //*************************************************
cin.get(); //*************************************************
}
}
void board(/*int scoreP1, int scoreP2*/){
cout << "\n\n\t\t P1 [" << scoreP1 << "]" << " P2 [" << scoreP2 << "]"; // fix (undefined)
cout << "\n\n\n\n\n";
cout << "\t\t | | \t\t\tCell orienation:" << endl;
cout << "\t\t " << cell[1] << " | " << cell[2] << " | " << cell[3] << endl;
cout << "\t\t_____|_____|_____\t\t\t 1 2 3" << endl;
cout << "\t\t | | " << endl;
cout << "\t\t " << cell[4] << " | " << cell[5] << " | " << cell[6] << "\t\t\t\t 4 5 6" << endl;
cout << "\t\t_____|_____|_____" << endl;
cout << "\t\t | | \t\t\t 7 8 9" << endl;
cout << "\t\t " << cell[7] << " | " << cell[8] << " | " << cell[9] << endl;
cout << "\t\t | | " << endl;
}
/***************************************************
for function 'piecePlacer'
pMove== 0 - invalid move
***************************************************/
抱歉,我的程序中有任何坏习惯,我对这方面还比较陌生。只是些样品。再次,我的问题是为什么这两个一起使用时,导致程序暂停?
2条答案
按热度按时间cngwdvgl1#
ignore()表示ignore是cin流的成员函数
阅读a little documentation,它有原型
该函数具有默认参数,由于您没有指定新参数,因此它将忽略输入中的一个字符。然后下一个输入将被cin.get()捕获
但是,cin.get()并没有指定它应该对输入做什么,所以您应该将其写成
cin.get(myvar);
或myvar = cin.get();
。也许递归调用piecePlacer与新的输入?dy2hfwbg2#
我想另一个答案并没有回答“为什么程序会暂停”,所以让我们继续。
cin.ignore()
使用默认参数count=1和delim=EOF,因此下一次击键将被忽略,即使它是EOF。cin.get()
是暂停程序,使其等待输入。这个输入(通常是指下一次击键),当它到来时,将被立即忽略,程序将继续。
例如:
上面的这个将暂停,等待输入,当它到来时,将返回/结束程序。