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c++ - tictactoe 游戏没有运算符 "=="匹配这些操作数

转载 作者:行者123 更新时间:2023-11-28 08:01:15 26 4
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这可能是非常低效和困惑的,但这是我的问题 - 它说的是“zz == defo && zo == defo”之类的东西等号用红色下划线表示错误:没有运算符“==”匹配这些操作数当我运行时,我收到一个错误,该错误太长而无法在此处发布- http://pastebin.com/KTEM0MZK

我做错了什么?提前致谢

这是我的代码-

#include <iostream>
#include <cstdlib> // for rand() and srand()
#include <ctime> // for time()
using namespace std;

string random1(){
srand(((time(0) - 23) * time(0)) - (9*time(0)));
if (rand() % 2 == 0){
return "1";
} else {
return "0";
}
}
string random2(){
srand((((time(0) - 89) * time(0)) - (9*time(0)) / 3) - 99);
if (rand() % 2 == 0){
return "1";
} else {
return "0";
}
}
int main(){
string zz = "-";
string zo = "-";
string zt = "-";
string oz = "-";
string oo = "-";
string ot = "-";
string tz = "-";
string to = "-";
string tt = "-";
for(int a = 1;a<=9;a++){
srand ( time(NULL) - 8 );
int ran1 = rand() % 3;
srand ( time(NULL) * 2);
int ran2 = rand() % 3;
int tote = (ran1 * 10) + ran2;
cout << endl << "format- 0 = zero, 1 = one, 2 = two" << endl;
int input;
cin >> input;
if(input == 00){
zz = "X";
} else if(input == 01){
zo = "X";
} else if(input == 02){
zt = "X";
} else if(input == 10){
oz = "X";
} else if(input == 11){
oo = "X";
} else if(input == 12){
ot = "X";
} else if(input == 20){
tz = "X";
} else if(input == 21){
to = "X";
} else if(input == 22){
tt = "X";
}
// now for the computers part
if(tote == 00){
zz = "O";
} else if(tote == 01){
zo = "O";
} else if(tote == 02){
zt = "O";
} else if(tote == 10){
oz = "O";
} else if(tote == 11){
oo = "O";
} else if(tote == 12){
ot = "O";
} else if(tote == 20){
tz = "O";
} else if(tote == 21){
to = "O";
} else if(tote == 22){
tt = "O";
}
printf ("|%d|%d|%d", zz, zo, zt);
cout << endl;
printf ("|%d|%d|%d", oz, oo, ot);
cout << endl;
printf ("|%d|%d|%d", tz, to, tt);

if(zz == "X" && zo == "X" && zt == "X" || oz == "X" && oo == "X" && ot == "X" || tz == "X" && to == "X" && tt == "X" || zz == "X" && oz == "X" && tz == "X" || zo == "X" && oo == "X" && to == "X" || zt == "X" && ot == "X" && tt == "X" || zz == "X" && oo == "X" && tt == "X"){
cout << endl << "X WINNER";
break;
}
if(zz == "O" && zo == "O" && zt == "O" || oz == "O" && oo == "O" && ot == "O" || tz == "O" && to == "O" && tt == "O" || zz == "O" && oz == "O" && tz == "O" || zo == "O" && oo == "O" && to == "O" || zt == "O" && ot == "O" && tt == "O" || zz == "O" && oo == "O" && tt == "O"){
cout << endl << "O WINNER";
break;
}
}
system("pause");
return 0;
}

最佳答案

您需要 #include <string> , 其中 operator== 已定义。

并且,正如 jrok 所述,你正在传递 std::stringprintf() : 使用 cout相反(因为你已经在使用)。

关于c++ - tictactoe 游戏没有运算符 "=="匹配这些操作数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11417494/

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