是否繼續(y/n)在此c代碼中不起作用?我希望它在輸入'y'時要求輸入一個字符串,如果輸入n,則退出程序。我嘗試了很多選擇,但無濟於事。 感謝您的幫助是否繼續(y/n)問題
do
{
i = 0, final = 0, s = 0;
printf("\n\nEnter Input String.. ");
scanf("%s", string);
while (string[i] != '\0')
if ((s = check(string[i++], s)) < 0)
break;
for (i = 0 ; i < nfinals ; i++)
if (f[i] == s)
final = 1;
if (final == 1)
printf("\n String is accepted");
else
printf("String is rejected");
printf("\nDo you want to continue.? \n(y/n) ");
}
while (getch() == 'y');
return getch();
}
前,應fflush你的緩衝區_doesn't WORK_請再具體些。 – 2014-10-19 03:57:31