又係c loop既問題

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>

int main(void)
{
int num;
char anyChar ='a';

for(num = 0; num <5; num++)
{
         printf("\nType in a letter form 'a' to 'e':\n");

         while ((anyChar=getche()) !='d')
         {
               printf("\nSorry , %c is incorrect.\n", anyChar);
               printf("Try again.\n");
               }
               printf("\nThat is it!\n");
               }
               printf("Game is over!\n");
               system("pause");
               }


點先可以限制輸入5次,5次後一定要出game is over(不論對錯)

0

0

TKY

進階會員

Rank: 3 Rank: 3 Rank: 3

2^#

發表於 2008-11-12 00:19 | 只看該作者

原帖由 cghncghn 於 2008-11-11 19:25 發表
int main(void)
{
int num;
char anyChar ='a';

for(num = 0; num <5; num++)
{
         printf("\nType in a letter form 'a' to 'e':\n");

         while ((anyChar=getche()) !='d')
         {
               printf("\nSorry , %c is incorrect.\n", anyChar);
               printf("Try again.\n");
         }
         printf("\nThat is it!\n");
}
printf("Game is over!\n");
system("pause");
}

你唔應該用for loop，唔係ge話，照你e+個program，要搭o岩5次要出到printf("Game is over!\n");呢句……
你應該係個while到加多個condition「num<=5」（入while loop前一開始set num做1），while入要就「num++」……

PS 你個program D括號最好對齊番位，唔係會好難睇……

TOP

梁炳

進階會員

Rank: 3 Rank: 3 Rank: 3

3^#

發表於 2008-11-12 00:33 | 只看該作者

睇下岩唔岩用

int main(void)
{
int count = 0;
char anyChar ='a';
printf("\nType in a letter form 'a' to 'e':\n");
while (anyChar=getche() || ++count <5)
{
if(anyChar != 'd')
printf("\nSorry , %c is incorrect.\nTry again.\n", anyChar);
else
{
printf("\nThat is it!\n");
break;
}
}
printf("Game is over!\n");
system("pause");
}

複製代碼

[ 本帖最後由梁炳於 2008-11-12 00:35 編輯 ]

TOP

cghncghn

中級會員

Rank: 2 Rank: 2

4^#

發表於 2008-11-12 00:52 | 只看該作者

原帖由梁炳於 2008-11-12 00:33 發表
睇下岩唔岩用int main(void)
{
int count = 0;
char anyChar ='a';

printf("\nType in a letter form 'a' to 'e':\n");
while (anyChar=getche() || ++count