扫描数字可能有问题 - C.

时间:2011-04-14 15:12:43

标签: c

  

可能重复:
  Program not doing what it should - C

您好,

以下程序只读取输入中的数字,并在违反冲击时停止,但一个大问题是它不会停止读取数字,更糟糕的是,不要在屏幕上打印它应该是什么。

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。

1 个答案:

答案 0 :(得分:3)

作为第一个提示,scanf返回读取的项目数,因此只有在读取48个项目(ASCII值为0)时才会触发条件(scanf("%d",&temp) ) != '0'。这种格式说明符不会发生这种情况,这就是你得到循环的原因。