Question

在C中使用简单的链接列表实现，如何告诉Splint我转移了data的所有权？

typedef struct {
    void* data;
    /*@null@*/ void* next;
} list;

static /*@null@*/ list* new_list(/*@notnull@*/ void* data)
{
    list* l;

    l = malloc(sizeof(list));

    if (l == NULL)
        return NULL;

    l->next = NULL;
    l->data = data;

    return l;
}

我收到此错误消息：

Implicitly temp storage data assigned to implicitly
                             only: list->data = data
  Temp storage (associated with a formal parameter) is transferred to a
  non-temporary reference. The storage may be released or new aliases created.
  (Use -temptrans to inhibit warning)

我想告诉Splint释放data的责任转移到列表数据结构。

Answer 1

解决方案位于function interfaces的Splint手册中。基本上，将函数签名更改为：

static /*@null@*/ list* new_list(/*@notnull@*/ /*@only@*/ void* data)
    /*@defines result->data @*/

虽然我们在执行此操作时会收到新错误：

int main()
{
    list* l = new_list("hej");

    return 0;
}


 Observer storage passed as only param:
                              new_list ("hej")
  Observer storage is transferred to a non-observer reference. (Use
  -observertrans to inhibit warning)

转移Splint中的存储所有权

1 个答案: