问题描述
#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->下;
while（tmp2！= NULL）
{
tmp1 = tmp2;
tmp2 = tmp1->下;
}
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->下一个;
}
return;
}
int main（int argc，char * argv []）
{
ls * head;
ls * tail;
int n，x;
head =（ls *）NULL;
tail =（ls *）NULL;
printf（\\\
1。add\\\
2。show\\\
3。exit\\\
）;
scanf（％d，& x）;
switch（x）
{
case 1：
scanf（％d，& n）;
add（* head，* tail，n）;
休息;
案例2：
show（* head，* tail）;
休息;
案例3：
返回0;
默认值：
break;
}
返回0;
}
当我使用gcc进行编译时
gcc -o lab5.out -Wall -pedantic lab5.c
$
lab5.c：在函数'main'中：
lab5 .c：84：3：error：'add'参数1的不兼容类型
lab5.c：16：6：note：预期'struct ls **'但参数类型为'ls'
lab5.c：84：3：error：'add'参数2的不兼容类型
lab5.c：16：6：注意：期望'struct ls **'但参数类型为'ls'
lab5.c：88：3：error：'show'参数1的不兼容类型
lab5.c：52：6：note：expected'struct ls **'但参数类型为'ls '
lab5.c：88：3：error：'show'参数2的不兼容类型
lab5.c：52：6：note：expected'struct ls **'但参数类型'ls'
对我而言，一切都很好...
参数类型是 ls ** 而不是 ls 就像编译器说的那样。
有人看到什么可能是错误的？
PS。我知道没有必要给 * tail 作为参数，但它是未使用的，因为我想开发这个'程序'...
& 代替的 * 会给你你想要的。以下是一些解释，以帮助您记忆。 * de - 引用它所附带的内容，意味着它将变量当作一个指针并给出该地址的值。
& 传递引用，这意味着它给出了该值存储的地址，或者更确切地说，将一个指针传递给变量。
既然你想传递一个指向变量头部和尾部的指针，你想把它们作为& head 和& tail ，它给你在另一端的值 **头和 **尾。在你习惯之前，它似乎是反直觉的。
I'm writing a list in C.Below is the source:
#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;
}
When I compile it with gcc
gcc -o lab5.out -Wall -pedantic lab5.c
I get strange errors:
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’
For me everything is OK...
And argument type is ls** and not ls as compiler says.
Someone see what can be wrong?
PS. I know that it's unnecessary to give *tail as argument and it's unused, however it will be, because I want to develop this 'program'...
As Daniel said in his comment and codaddict said in his answer, using & instead of * will give you what you want. Here's a bit of an explanation though, to help you remember.
* de-references what it's attached to, meaning it takes the variable as if it's a pointer and gives the value at that address.
& passes the reference, meaning it gives the address that the value is stored at, or rather, passes a pointer to the variable.
Since you want to pass a pointer to the variables head and tail, you want to pass them as &head and &tail, which gives you the values **head and **tail at the other end. It seems counter-intuitive until you get used to it.
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