0
你好,
下面的程序只能读取输入的数字,当鲁尔房屋被侵犯别停下来,但1大问题是,它不会停止阅读数字,更糟的是,不要在屏幕上打印它应该如何。
代码:
#include <stdio.h>
#include <string.h>
void SIFT(int x_arr[ ], int y_arr[]);
int main()
{
int x[20] = {0} , y[20] = {0};
int m=0,temp=0,curr=0,i=0,j=0;
printf("Please enter your numbers now:\n\n");
/*enter numbers one by one. if x[i+1] value < x[i] value, err msg.
when user want to end the series he must enter '0' which means end of string (it wont included in x[]) */
while ((scanf("%d",&temp)) != '0')
{
if (temp >= curr)
{
x[i] = temp;
curr = temp;
i++;
}
else
{
printf("The numbers are not at the right order !\n\nProgram will now terminate...\n\n");
}
}
SIFT(x,y);
for (i=0 ; y[i]=='0' ; i++) /*strlen(y) without ('0')'s includes*/
m++;
/*Prints m , y's organs*/
printf("\n\nm = %d",m);
printf("Y = ");
while (y[j]!='0')
{
printf ("%d ,",y[j]);
j++;
}
return 0;
}
void SIFT(int x_arr[ ], int y_arr[])
{
int i=0,j=0;
while (x_arr[i] != '0')
{
if (x_arr[i] == x_arr[i+1]) /*if current val. equals next val. -> jump dbl at x_arr*/
{
y_arr[j] = x_arr[i];
i+=2;
j++;
}
else
{
y_arr[j]=x_arr[i];
i++;
j++;
}
}
}
请帮我解决这个问题... 日Thnx。
这不是以前的问题,尽管它几乎是相同的代码。 – Mat 2011-04-14 15:17:50
'scanf(“%d”,&temp)'“有2个结果”。一个这样的“结果”是函数的返回值:1如果一切正常;否则为0或EOF。另一个“结果”(当返回1时)是变量temp的值(用户输入的值)。你需要检查“结果”...(如非常伪代码)if(scanf()== 1){if(temp == 0)...; }' – pmg 2011-04-14 15:38:47