我写在C 列表下面是源:错误:不兼容的类型参数
#include <stdio.h>
#include <stdlib.h>
struct list {
int value;
struct list *next;
};
typedef struct list ls;
void add (ls **head, ls **tail, int val)
{
ls *new, *tmp1, *tmp2;
if (NULL == *head)
{
new = (ls*)malloc(sizeof(ls));
*head = new;
*tail = new;
new->value = val;
new->next = NULL;
return;
}
else
{
tmp1 = *head;
tmp2 = tmp1->next;
while (tmp2 != NULL)
{
tmp1 = tmp2;
tmp2 = tmp1->next;
}
new = (ls*)malloc(sizeof(ls));
new->value = val;
new->next = NULL;
*tail = new;
return;
}
}
void show (ls **head, ls **tail)
{
int i;
ls *tmp;
while (tmp->next != NULL)
{
printf("%d: %d", i, tmp->value);
i++;
tmp=tmp->next;
}
return;
}
int main (int argc, char *argv[])
{
ls *head;
ls *tail;
int n, x;
head = (ls*)NULL;
tail = (ls*)NULL;
printf("\n1. add\n2. show\n3. exit\n");
scanf("%d", &x);
switch (x)
{
case 1:
scanf("%d", &n);
add(*head, *tail, n);
break;
case 2:
show(*head, *tail);
break;
case 3:
return 0;
default:
break;
}
return 0;
}
当我用gcc
gcc -o lab5.out -Wall -pedantic lab5.c
编译它,我得到奇怪的错误:
lab5.c: In function ‘main’:
lab5.c:84:3: error: incompatible type for argument 1 of ‘add’
lab5.c:16:6: note: expected ‘struct ls **’ but argument is of type ‘ls’
lab5.c:84:3: error: incompatible type for argument 2 of ‘add’
lab5.c:16:6: note: expected ‘struct ls **’ but argument is of type ‘ls’
lab5.c:88:3: error: incompatible type for argument 1 of ‘show’
lab5.c:52:6: note: expected ‘struct ls **’ but argument is of type ‘ls’
lab5.c:88:3: error: incompatible type for argument 2 of ‘show’
lab5.c:52:6: note: expected ‘struct ls **’ but argument is of type ‘ls’
对我来说,一切都OK了...
参数类型为ls**
,而不是编译器说的ls
。
有人看到什么可能是错的?
PS。我知道这是没有必要给*tail
作为参数,它是未使用的,但是这将是的,因为我想开发这个“计划” ......
你需要''而不是'*'在通话中。 – 2012-07-18 16:33:03
当然......它解决了这个问题。非常感谢... – kostek 2012-07-18 16:39:43