错误：参数的不兼容类型 [英] error: incompatible type for argument

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'...

解决方案

正如Daniel在他的评论中所说的，codaddict在他的回答中表示，用& 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'...

解决方案

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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