C语言 从不兼容的指针类型警告传递参数
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Passing argument from incompatible pointer type warning
提问by hora
I've been trying to figure out pointers in C most of today, even asked a questionearlier, but now I'm stuck on something else. I've got the following code:
我今天大部分时间都在试图找出 C 中的指针，甚至早些时候问过一个问题，但现在我被困在别的事情上了。我有以下代码：
typedef struct listnode *Node;
typedef struct listnode {
void *data;
Node next;
Node previous;
} Listnode;
typedef struct listhead *LIST;
typedef struct listhead {
int size;
Node first;
Node last;
Node current;
} Listhead;
#define MAXLISTS 50
static Listhead headpool[MAXLISTS];
static Listhead *headpoolp = headpool;
#define MAXNODES 1000
static Listnode nodepool[MAXNODES];
static Listnode *nodepoolp = nodepool;
LIST *ListCreate()
{
if(headpool + MAXLISTS - headpoolp >= 1)
{
headpoolp->size = 0;
headpoolp->first = NULL;
headpoolp->last = NULL;
headpoolp->current = NULL;
headpoolp++;
return &headpoolp-1; /* reference to old pointer */
}else
return NULL;
}
int ListCount(LIST list)
{
return list->size;
}
Now in a new file I have:
现在在一个新文件中我有：
#include <stdio.h>
#include "the above file"
main()
{
/* Make a new LIST */
LIST *newlist;
newlist = ListCreate();
int i = ListCount(newlist);
printf("%d\n", i);
}
When I compile, I get the following warning (the printfstatement prints what it should):
当我编译时，我收到以下警告（该printf语句打印出它应该的内容）：
file.c:9: warning: passing argument 1 of ‘ListCount' from incompatible pointer type
Should I be worried about this warning? The code seems to do what I want it to, but I'm obviously very confused about pointers in C. After browsing questions on this site, I found that if I make the argument to ListCount (void *) newlist, I don't get the warning, and I don't understand why, nor what (void *)really does...
我应该担心这个警告吗？代码似乎按照我的意愿行事(void *) newlist，但我显然对 C 中的指针感到非常困惑。在本网站上浏览问题后，我发现如果我向 ListCount 提出论点，我不会收到警告，并且我不明白为什么，也不明白什么是(void *)真的......
Any help would be appreciated, thanks.
任何帮助将不胜感激，谢谢。
回答by Alok Singhal
You're getting confused because of multiple typedefs. LISTis a type representing a pointer to struct listhead. So, you want your ListCreatefunction to return a LIST, not a LIST *:
由于多个 typedef，您会感到困惑。 LIST是表示指向 的指针的类型struct listhead。因此，您希望ListCreate函数返回 a LIST，而不是 a LIST *：
LIST ListCreate(void)
The above says: ListCreate()function will return a pointer to a new list's head if it can.
上面说：ListCreate()如果可以，函数将返回一个指向新列表头部的指针。
Then you need to change the returnstatement in the function definition from return &headpoolp-1;to return headpoolp-1;. This is because you want to return the last available head pointer, and you have just incremented headpoolp. So now you want to subtract 1 from it and return that.
然后您需要return将函数定义中的语句从return &headpoolp-1;更改为return headpoolp-1;。这是因为您想返回最后一个可用的头指针，而您刚刚增加了headpoolp. 所以现在你想从中减去 1 并返回它。
Finally, your main()needs to be update to reflect the above changes:
最后，您main()需要更新以反映上述更改：
int main(void)
{
/* Make a new LIST */
LIST newlist; /* a pointer */
newlist = ListCreate();
int i = ListCount(newlist);
printf("%d\n", i);
return 0;
}
