我正在编写此代码进行训练,我遇到了一个问题,如果我的用户写他的名字后跟一个空格和其他东西,程序会搞砸我的流程。因此,如果您尝试使用小程序,当它询问名称时,请像"Robert Red"一样输入,这会更容易。当您在空格后放置其他内容时,就会出现问题,如果您只输入"罗伯特",一切都会很好。
这是代码:
// Description: This is a simple replica of the Japanese game Rock, Paper and
// Scissors.
// Author: Ernesto Campese
// Last Update: 11/04/2018
// Version: 0.0.1
#include "std_lib_facilities.h"
int main() {
string username = "";
char userinput;
int rounds = 0;
int wins = 0;
int draws = 0;
int loses = 0;
int user_secret = 0;
vector<string> options = {"Paper", "Scissors", "Rock"};
cout << "Enter your name: ";
cin >> username;
cout << "Welcome " << username << ", this is the game of Rock, Paper and Scissors.n";
cout << username << " how many rounds you want to do? ";
cin >> rounds;
if (rounds <= 0) {
cout << "You need to play at least one round!n";
rounds++;
}
cout << "The game is based on " << rounds << " rounds, you versus the CPU.n";
cout << "Are you ready? (y/n): ";
cin >> userinput;
if (userinput != 'y') {
cout << "nThank you.nProgram Terminated by " << username;
return 0;
}
for(int i = 1; i <= rounds; i++) {
// Title of the rounds
if (i == 1) {
cout << "nLet's start the first round!n";
} else {
cout << "Round n. " << i << " begins!n";
}
// USER makes a move
cout << "Which is your move? (r,p,s): ";
cin >> userinput;
cout << 'n' << username << " says... ";
switch (userinput) {
case 'r':
cout << "Rockn";
user_secret = 2;
break;
case 'p':
cout << "Papern";
user_secret = 0;
break;
case 's':
cout << "Scissorsn";
user_secret = 1;
break;
default:
cout << "something weird...n";
break;
}
// CPU makes a move
int cpu_secret = rand() % 3;
cout << "CPU says... " << options[cpu_secret] << "!n";
// The program calculates the result.
if (user_secret == cpu_secret) {
draws++;
cout << username << " and the CPU draws!nn";
} else if (user_secret == 0 && cpu_secret == 2) {
wins++;
cout << username << " wins!nn";
} else if (user_secret == 1 && cpu_secret == 0) {
wins++;
cout << username << " wins!nn";
} else if (user_secret == 2 && cpu_secret == 1) {
wins++;
cout << username << " wins!nn";
} else {
loses++;
cout << username << " lose!nn";
}
}
cout << "nnBattle End!n";
if (wins > loses) {
cout << username << " won the battle!n";
} else if (loses > wins) {
cout << username << " lost the battle!n";
} else {
cout << username << " draws the battle!n";
}
cout << "Thank you " << username << "!n";
}
你可以在这里试试:试试我谢谢!
operator>>
在找到空格字符时停止读取输入。
使用 std::getline()
读取带有空格的用户输入。
使用代码的示例:
cout << "Enter your name: ";
getline(cin, username);
如果您希望用户能够键入包含空格的名称,请使用 std::getline()
而不是 operator>>
:
getline(cin, username);
否则,如果希望用户只输入 1 个单词作为名称,并且希望忽略用户可能输入的任何其他内容,请使用 std::cin.ignore()
:
#include <limits>
...
cin >> username;
cin.ignore(numeric_limits<streamsize>::max(), 'n');
或者,您可以使用 std::getline()
读取一行,然后将 std::istringstream
与 operator>>
一起使用以提取该行的第一个单词:
#include <sstream>
...
string line;
getline(cin, line);
istringstream(line) >> username;