用循环C语言程序问题 [英] C Program Problems with Loops
问题描述
我很新的编程和目前工作的一个程序(它没有在附近完成的),但我想不出如何以及为什么这是不行的。我必须失去了与'如果'和'其他'
东西 的#include<&stdio.h中GT;INT主要(无效)
{
的printf(欢迎来到机器人GAME!\\ n
请输入您的出发点列保存号码一到三的范围\\ n。); 诠释A,B;
如果(1 << = A&下; = 3);
{
scanf函数(%i的,&安培;一);
}
其他
{
的printf(对不起再试一次。);
} 的printf(请输入您的出发点的行保留号码一到三的范围\\ n。);
如果(1&GT; = b将; = 3);
scanf函数(%i的,和b);
的printf(什么是你的下一步行动?选择\\ n
1.右键\\ n
2.Left \\ n
3.Up \\ n
4.Down \\ n);}
第一个臭虫
INT A,B;
如果(1 << = A&下; = 3);
您如何检查 A
在如果
的条件,甚至是未初始化?首先通过初始化值 A
和 B
(用 1 $ C $对其进行修复C>)然后去解决二错误
第二个错误
如果(1 << = A&LT; = 3);
删除;
在如果
条件,然后去修复第三错误
第三个错误
如果(1 << = A&LT; = 3);
是不是该做你想要什么。它应该
如果(A&GT; = 1&安培;&安培; A&LT; = 3)
现在同样适用于如果(1&GT; = B&LT; = 3);
按照每个上面的步骤来解决这个问题。
第四错误
您有 INT主要(无效)
。因此需要返回0
在结束主
I'm very new to programming and currently working on a program (it is no where near done) but I cannot figure out how and why this won't work. I must be missing something with the 'if' and 'else'
#include <stdio.h>
int main(void)
{
printf("Welcome to the ROBOT GAME!\n"
"Please enter the column of your starting point. Keep numbers in the range of one to three.\n");
int a, b;
if(1<=a<=3);
{
scanf("%i", &a);
}
else
{
printf("Sorry try again.");
}
printf("Please enter the row of your starting point. Keep numbers in the range of one to three.\n");
if(1>=b<=3);
scanf("%i", &b);
printf("What is your next move? Pick\n"
"1.Right\n"
"2.Left\n"
"3.Up\n"
"4.Down\n");
}
First Bug
int a, b;
if(1<=a<=3);
How can you check a
in if
condition even it not initialized?. First Fix it by initializing value to a
and b
( with 1
) Then go for to fix Second Bug
Second Bug
if(1<=a<=3);
Remove ;
after if
condition then go for Fix Third Bug
Third Bug
if(1<=a<=3);
is not suppose to do what you want. it should
if(a>= 1 && a <= 3)
Now same is true for if(1>=b<=3);
Follow every above steps to fix this.
Fourth Bug
You have int main(void)
. So need to return 0
at end of main
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