为什么不能用函数读入结构指针

[AnAzureAssassin]

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<time.h>
struct list{
int N;
};
typedef struct list *List;
void Read_Sequence(List sequence){
scanf("%d",&sequence->N);//这里是错误的
return;
}
int main(void) {
List sequence;
Read_Sequence(sequence);
return 0;
}
//为什么标注的地方是错误的

[WNs_ACE]

sequence对象都没有初始化，需要初始化指针的啊.....
List sequence = new list;
下面是我测试代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<time.h>
struct list
{
	int N;
};

typedef struct list *List;

void Read_Sequence(List sequence)
{
	scanf("%d",&(sequence->N));
	return;
}
int main(void)
 {
	List sequence = new list();
	Read_Sequence(sequence);
	
	printf("N:%d\n", sequence->N);
	
	
	return 0;
}

[WNs_ACE]

重新补一下代码，忘记写释放指针了

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<ctype.h>
#include<time.h>
struct list
{
	int N;
};

typedef struct list *List;

void Read_Sequence(List sequence)
{
	scanf("%d",&(sequence->N));//这里是错误的
	return;
}
int main(void)
 {
	List sequence = new list();
	Read_Sequence(sequence);
	
	printf("N:%d\n", sequence->N);
	
	delete sequence;
	return 0;
}