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问题描述
使用在C简单的链表实现，我怎么告诉夹板，我转数据的所有权？
Using a simple linked list implementation in C, how do I tell Splint that I am transfer ownership of data?
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;
}
我收到此错误信息：
I get this error message:
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)
我要告诉夹板固定的释放数据的责任被转移到列表中的数据结构。
I want to tell Splint that responsibility of freeing data is transfered to the list data-structure.
推荐答案
解决方案是在的的。基本上，函数签名改成这样：
The solution is in the Splint manual for function interfaces. Basically, change the function signature to this:
static /*@null@*/ list* new_list(/*@notnull@*/ /*@only@*/ void* data)
/*@defines result->data @*/
虽然这样做的时候，我们会得到一个新的错误：
Although we'll get a new error when doing this:
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)
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