如何修复：从不兼容的指针类型初始化

Question

当我编译和运行程序时，它说：在函数 'addToTheEnd' 中：[警告] 从不兼容的指针类型初始化 // 它指向这一行 -> knot_t *当前=开始；

如何解决？我是 C 的新手，所以我不明白应该改变什么。我试图理解它，但我做不到。我的目标是在我运行它时从该程序中获取一些输出，但没有任何显示。

#include <stdio.h>
#include <stdlib.h>

typedef struct knot {
    int number;
    struct knot *next;
} knot_t;

int addToTheEnd(knot_t **start, int value) {
    knot_t *new_ = (knot_t *)malloc(sizeof(knot_t));
    if (new_ == NULL) {
        return 1;
    }
    new_ -> number = value;
    new_ -> next = NULL;
   
    if (start == NULL) {
        *start = new_;
    } else {
        knot_t *current = start;
        while (current -> next != NULL) {
            current = current -> next;
        }
        current -> next = new_;
    }
   
    return 0;
}

void printList(knot_t *start) {
    knot_t *current = start;
    while (current != NULL) {
        printf("%d ", current->number);
        current = current -> next;
    }
}

void clearList(knot_t *start) {
    knot_t *current = start;
    while (current != NULL) {
        knot_t* trenutni = current;
        current = current -> next;
        free(trenutni);
    }
}

int main() {
    knot_t *start = NULL;
    int i = 0;
    for (i = 0; i < 10; i++) {
        if(addToTheEnd(&start, i)) {
        printf("Fail\n");
return EXIT_FAILURE;
}
    }
   
    printList(start);
    clearList(start);
   
    return EXIT_SUCCESS;
}

Answer 1

该函数具有 knot_t ** 类型的参数 start。

int addToTheEnd(knot_t **start, int value) {

而局部变量 current 的类型为 knot_t *。

knot_t *current = start;

因此，由于没有从类型 knot_t ** 到类型 knot_t * 的隐式转换，编译器会发出错误。

你的意思好像是

knot_t *current = *start;

也是if语句的条件

 if (start == NULL) {
        *start = new_;

必须是

 if (*start == NULL) {
        *start = new_;

另外，当函数在成功的情况下 return 1 而不是 0 时，逻辑上会更加一致。

函数可以这样定义

int addToTheEnd( knot_t **start, int value ) 
{
    knot_t *new_knot = malloc( sizeof( knot_t ) );
    int success = new_knot != NULL;

    if ( success )
    {
        new_knot -> number = value;
        new_knot -> next = NULL;
   
        while ( *start != NULL ) start = &( *start )->next;

        *start = new_knot;
    } 
   
    return success;
}

并且由于函数 printList 不会更改指向的节点，因此至少应声明为

void printList( const knot_t *start) {
    const knot_t *current = start;
    while (current != NULL) {
        printf("%d ", current->number);
        current = current -> next;
    }

Answer 2

这对我有用：

int addToTheEnd(knot_t** start, int value) {
  knot_t* new_ = (knot_t*)malloc(sizeof(knot_t));
  if (new_ == NULL) {
    return 1;
  }
  new_->number = value;
  new_->next = NULL;

  if (*start == NULL) {    // start -> *start
    *start = new_;
  }
  else {
    knot_t* current = *start;    // start -> *start
    while (current->next != NULL) {
      current = current->next;
    }
    current->next = new_;
  }

  return 0;
}

如何修复：从不兼容的指针类型初始化

How to fix: initialization from incompatible pointer type

c

struct

linked-list

implicit-conversion

singly-linked-list