顺序结构存储的串删除子串,该如何解决

顺序结构存储的串删除子串
#include "stdafx.h"
#include "iostream.h"
#define maxsize 6
typedef struct
{
char ch[maxsize];
int len;
}seqstring;

seqstring *WEEPSTR(seqstring *S,int i,int j)
{
int k,l;
if(i+j>S->len) printf("越界\n");
else
{
for(k=0;k<j;k++)
for(l=i;l<i+j-k;l++)
S->ch[i]=S->ch[i+1];
}
return (S);
}

int main()
{
int m,n,i;
seqstring *S;
seqstring *WEEPSTR(seqstring *S, int i,int j);
puts("please put in a string with the length of 6:");
for(i=0;i<6;i++)
S->ch[i]=getchar();
printf("start from:");
scanf("%d",&m);
printf("the length you want to weep:");
scanf("%d",&n);
WEEPSTR(S, m,n );
puts("the new string is: ");
puts(S->ch);
return 0;
}
有一个警告，运行以后出现内存不能写入，请各位路过的大神帮忙看一下~~万谢~~~~~

------解决方案--------------------
楼主，S没分配空间：

S = (seqstring *) malloc (sizeof(seqstring));
------解决方案--------------------
参考:

C/C++ code


#include <stdio.h>
#include <stdlib.h>
#define maxsize 6
typedef struct  
{
    char ch[maxsize];
    int len;
}seqstring;
seqstring *WEEPSTR(seqstring *S,int i,int j)
{
    int k,l;  
    if(i+j>S->len) printf("越界\n");
    else  
    {
    /*    for(k=0;k<j;k++)
            for(l=i;l<i+j-k;l++)
                S->ch[i]=S->ch[i+1];
                */
        for(k=i+j; k<S->len;++k)
        {
            S->ch[i++] = S->ch[k];
        }
        S->ch[i]='\0';
    }
    return (S);
}

int main()
{
    int m,n,i; 
    seqstring *S;
    seqstring *WEEPSTR(seqstring *S, int i,int j);
    S = (seqstring *) malloc (sizeof(seqstring)); //分配空间
    puts("please put in a string with the length of 6:");
    S->len = 6; // 对长度赋值
/*    for(i=0;i<6;i++)
        S->ch[i]=getchar();
        */
    scanf("%s", S->ch); //用scanf()不要管回车

    printf("start from:");
    scanf("%d",&m);
    printf("the length you want to weep:");
    scanf("%d",&n);
    WEEPSTR(S, m,n );
    puts("the new string is: ");
    puts(S->ch);
    return 0;
}

顺序结构存储的串删除子串,该如何解决

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