为什么这个指向结构的指针没有指向正确的结构?
Why does this pointer to a struct not point to the correct one?
我正在构建一个 NFA 并希望通过创建指向其他 State
的 State
结构来实现。 NFA 构造过程要求我跟踪哪些 State
指向 NULL
,然后在我知道它们应该指向什么 State
时修补它们。
但是当我更新链表时,它并没有更新指针State
。我想我没有正确引用和更新 NULL
指针。
这是有问题的代码的简化版本:
#include <stdio.h>
#include <stdlib.h>
typedef struct State State;
struct State
{
char c;
State *out;
};
typedef struct List List;
struct List
{
State *s;
// has a next member that is irrelevant here.
};
State *State_new(char c, State *out)
{
State *s;
s = malloc(sizeof(*s));
s->c = c;
s->out = out;
return s;
}
void *List_new(State **outpp)
{
List *slist = malloc(sizeof(*slist));
/*
* Dereference the pointer to a pointer of a State
* to get a pointer to a state
*/
slist->s = *outpp;
return slist;
}
int main()
{
State *a = State_new('a', NULL);
List *l = List_new(&(a->out));
/* This printf() will result in a seg fault, since a->out is NULL. */
//printf("%c\n", a->out->c);
/* change what State struct is pointed to by l */
l->s = State_new('b', NULL);
/* why is this not b? */
//printf("%c\n", a->out->c);
return 0;
}
a->out->c
不是 'b'
因为您正在将指针的副本存储在 List 的成员中。你给了一个 State**
作为参数,但你也应该这样存储它。如果不是这种情况,您可以简单地发送一个 State *outp
并写成 slist->s = outp;
。
#include <stdio.h>
#include <stdlib.h>
typedef struct State State;
struct State
{
char c;
State *out;
};
typedef struct List List;
struct List
{
State **s; //<--- HERE
// has a next member that is irrelevant here.
};
State *State_new(char c, State *out)
{
State *s;
s = malloc(sizeof(*s));
s->c = c;
s->out = out;
return s;
}
void *List_new(State **outpp)
{
List *slist = malloc(sizeof(*slist));
/*
* Dereference the pointer to a pointer of a State
* to get a pointer to a state
*/
slist->s = outpp; //<<--- HERE
return slist;
}
int main()
{
State *a = State_new('a', NULL);
List *l = List_new(&(a->out));
/* This printf() will result in a seg fault, since a->out is NULL. */
//printf("%c\n", a->out->c);
/* change what State struct is pointed to by l */
*l->s = State_new('b', NULL);
printf("%c\n", a->out->c);
return 0;
}
我正在构建一个 NFA 并希望通过创建指向其他 State
的 State
结构来实现。 NFA 构造过程要求我跟踪哪些 State
指向 NULL
,然后在我知道它们应该指向什么 State
时修补它们。
但是当我更新链表时,它并没有更新指针State
。我想我没有正确引用和更新 NULL
指针。
这是有问题的代码的简化版本:
#include <stdio.h>
#include <stdlib.h>
typedef struct State State;
struct State
{
char c;
State *out;
};
typedef struct List List;
struct List
{
State *s;
// has a next member that is irrelevant here.
};
State *State_new(char c, State *out)
{
State *s;
s = malloc(sizeof(*s));
s->c = c;
s->out = out;
return s;
}
void *List_new(State **outpp)
{
List *slist = malloc(sizeof(*slist));
/*
* Dereference the pointer to a pointer of a State
* to get a pointer to a state
*/
slist->s = *outpp;
return slist;
}
int main()
{
State *a = State_new('a', NULL);
List *l = List_new(&(a->out));
/* This printf() will result in a seg fault, since a->out is NULL. */
//printf("%c\n", a->out->c);
/* change what State struct is pointed to by l */
l->s = State_new('b', NULL);
/* why is this not b? */
//printf("%c\n", a->out->c);
return 0;
}
a->out->c
不是 'b'
因为您正在将指针的副本存储在 List 的成员中。你给了一个 State**
作为参数,但你也应该这样存储它。如果不是这种情况,您可以简单地发送一个 State *outp
并写成 slist->s = outp;
。
#include <stdio.h>
#include <stdlib.h>
typedef struct State State;
struct State
{
char c;
State *out;
};
typedef struct List List;
struct List
{
State **s; //<--- HERE
// has a next member that is irrelevant here.
};
State *State_new(char c, State *out)
{
State *s;
s = malloc(sizeof(*s));
s->c = c;
s->out = out;
return s;
}
void *List_new(State **outpp)
{
List *slist = malloc(sizeof(*slist));
/*
* Dereference the pointer to a pointer of a State
* to get a pointer to a state
*/
slist->s = outpp; //<<--- HERE
return slist;
}
int main()
{
State *a = State_new('a', NULL);
List *l = List_new(&(a->out));
/* This printf() will result in a seg fault, since a->out is NULL. */
//printf("%c\n", a->out->c);
/* change what State struct is pointed to by l */
*l->s = State_new('b', NULL);
printf("%c\n", a->out->c);
return 0;
}