我正在为大学计算机编程课制作一个井字棋游戏。我有几行需要在程序中打印 3 次。我为此创建了一个函数，我调用了它并且没有收到任何错误，但问题是该函数没有运行。它只有几行要打印，但它没有这样做。请阅读源文件查看代码: C源:https://dl.dropboxusercontent.com/u/62524851/Untitled1.c 该程序的 Exe 文件:https://dl.dropboxusercontent.com/u/62524851/Untitled1.exe
#include <stdio.h>
#include <stdlib.h>
void text(char layout[9], int guide[9])
{
system("cls");
printf("\n\n\t\t ----------Tic tac toe----------");
printf("\n\nEnter number of the column you want to make your move in.\nNumbers of columns are shown below:\nPress 10 to quit.\n\n");
printf("\t\t\t%d\t%d\t%d", guide[0], guide[1], guide[2]);
printf("\n\n\t\t\t%d\t%d\t%d", guide[3], guide[4], guide[5]);
printf("\n\n\t\t\t%d\t%d\t%d\n\n", guide[6], guide[7], guide[8]);
printf("\t\t\t%c\t%c\t%c", layout[0], layout[1], layout[2]);
printf("\n\n\t\t\t%c\t%c\t%c", layout[3], layout[4], layout[5]);
printf("\n\n\t\t\t%c\t%c\t%c\n\n", layout[6], layout[7], layout[8]);
}
void main(void)
{
int guide[9] = {1, 2, 3, 4, 5, 6, 7, 8, 9}, user, user2, x, i;
char layout[9] = {'.', '.', '.', '.', '.', '.', '.', '.', '.'}, c, name2[8], a[8];
do
{
system("cls");
printf("\n\n\t\t ----------Tic tac toe----------");
printf("\n\t\t By Muhammad Saad Masroor. 23507");
printf("\n\n\n\n\n\n\n\n\n\n\n\nPress c to continue");
c = getch();
} while (c != 'c');
system("cls");
printf("\n\n\t\t ----------Tic tac toe----------\n\n");
printf("With whome do you want to play?\n\nA. Computer\nB. Friend\n");
c = getch();
puts("Enter your name: ");
gets(a);
if (c == 'a')
{
system("cls");
printf("\n\n\n\n\n\n\n\n\t\t\t Lets play!");
printf("\n\n\t\t Between %s and Computer", a);
getch();
}
else if (c == 'b')
{
puts("Enter your friend's name: ");
gets(name2);
system("cls");
printf("\n\n\n\n\n\n\n\n\t\t\t Lets play!");
printf("\n\n\t\t Between %s and %s", a, name2);
getch();
}
while (x != 10)
{
void text(char layout[9], int guide[9]);
printf("%s's turn", a);
scanf("%d", &user);
if (user == user2)
{
printf("Invalid input: %d, enter another number", user);
scanf("%d", &user);
}
switch (user)
{
case (1):
layout[0] = 'X';
break;
case (2):
layout[1] = 'X';
break;
case (3):
layout[2] = 'X';
break;
case (4):
layout[3] = 'X';
break;
case (5):
layout[4] = 'X';
break;
case (6):
layout[5] = 'X';
break;
case (7):
layout[6] = 'X';
break;
case (8):
layout[7] = 'X';
break;
case (9):
layout[8] = 'X';
break;
case (10):
x = 10;
printf("You quit");
break;
default:
printf("Invalid input");
getch();
break;
}
void text(char layout[9], int guide[9]);
// User wins
if (layout[0] == 'X' && layout[1] == 'X' && layout[2] == 'X' ||
layout[3] == 'X' && layout[4] == 'X' && layout[5] == 'X' ||
layout[6] == 'X' && layout[7] == 'X' && layout[8] == 'X' ||
layout[0] == 'X' && layout[3] == 'X' && layout[6] == 'X' ||
layout[1] == 'X' && layout[4] == 'X' && layout[7] == 'X' ||
layout[2] == 'X' && layout[5] == 'X' && layout[8] == 'X' ||
layout[0] == 'X' && layout[4] == 'X' && layout[8] == 'X' ||
layout[2] == 'X' && layout[4] == 'X' && layout[6] == 'X')
{
printf("%s wins!\n", a);
x = 10;
break;
}
else if (user == 10)
{
break;
}
else if (c == 'b')
{
printf("%s's turn", name2);
scanf("%d", &user2);
}
if (user2 == user)
{
printf("Invalid input: %d, enter another number", user);
scanf("%d", &user2);
}
if (user2 == 10)
{
printf("You quit");
getch();
x = 10;
}
else if (c == 'a')
{
user2 = rand()%(9);
if (user2 == user)
{
user2 = rand()%(9);
}
printf("Computer's turn: %d", user2);
}
switch (user2)
{
case (1):
layout[0] = 'O';
getch();
break;
case (2):
layout[1] = 'O';
break;
case (3):
layout[2] = 'O';
break;
case (4):
layout[3] = 'O';
break;
case (5):
layout[4] = 'O';
break;
case (6):
layout[5] = 'O';
break;
case (7):
layout[6] = 'O';
break;
case (8):
layout[7] = 'O';
break;
case (9):
layout[8] = 'O';
break;
default:
printf("Invalid input");
break;
}
void text(char layout[9], int guide[9]);
// Computer/2nd user wins
if (layout[0] == 'O' && layout[1] == 'O' && layout[2] == 'O' ||
layout[3] == 'O' && layout[4] == 'O' && layout[5] == 'O' ||
layout[6] == 'O' && layout[7] == 'O' && layout[8] == 'O' ||
layout[0] == 'O' && layout[3] == 'O' && layout[6] == 'O' ||
layout[1] == 'O' && layout[4] == 'O' && layout[7] == 'O' ||
layout[2] == 'O' && layout[5] == 'O' && layout[8] == 'O' ||
layout[0] == 'O' && layout[4] == 'O' && layout[8] == 'O' ||
layout[2] == 'O' && layout[4] == 'O' && layout[6] == 'O')
{
if (c == 'a')
{
printf("Computer wins!\n");
getch();
x = 10;
break;
}
else if (c == 'b')
{
printf("%s wins!", name2);
getch();
x = 10;
break;
}
}
}
}
与原始代码相比，代码重新缩进并稍微重新格式化。
最佳答案
在继续之前，这段代码:
switch (user)
{
case (1):
layout[0]='X';
break;
...
应替换为:
layout[user-1] = 'X';
在进行分配之前，您需要验证移动是否有效:
if (user == 10)
...report quit and break loop...
else if (user <= 0 || user > 10)
...report error and collect new number...
并且该模因被重复使用，使用 40 多行，其中 5 行就足够了。代码中还有其他明显的重复，最好避免。
<小时/>作为您真正的问题，问题在于您没有在任何地方调用函数 text() 。您刚刚声明了多次。
因此，例如，在 main() 中，您有:
while (x != 10)
{
void text(char layout[9], int guide[9]);
printf("%s's turn", a);
scanf("%d", &user);
这再次声明了 text()，但没有调用它。你需要这样的东西:
while (x != 10)
{
text(layout, guide);
printf("%s's turn", a);
scanf("%d", &user);
关于无法在 c 中打印我的函数中的任何内容，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19899990/