我无法使用链表打印多项式

Question

void create(struct node *head);                   //declaring functions
void display(struct node *head);     
struct node                                 //creating a struct datatype for nodes        
{
 int coeff;
 int power;
 struct node* next;
};
  struct node* poly1=NULL;               //head pointer for a sample polynomial

  void main()
 {
   int coeff,power,deg,i;
   clrscr();                                              //main function
   printf("\n enter polynomial 1 ");
   create(poly1);
   display(poly1);
   getch();
 }

  void create(struct node *head)
 {
  struct node *newnode,*temp;
  int exp,num,n,i;
  printf("\n enter no. of terms in your expression=");
  scanf("%d",&n);
  for(i=0;i<n;i++)
  {
   newnode=(struct node*)malloc(sizeof(newnode));
   newnode->next=NULL;
   printf("\n enter power=");
   scanf("%d",&exp);
   newnode->power=exp;
   printf("\n enter coefficient=");
   scanf("%d",&num);
   newnode->coeff=num;
    if(head==NULL)
     head=newnode;
    else
     {
      temp=head;
      while(temp->next!=NULL)
      {
       temp=temp->next;
      }
       temp->next=newnode;
     }
 }
}
 void display(struct node *head)
 {
  struct node *temp;
  temp=head;                                                 
  while(temp->next!=NULL)
  {
   printf("%dx^%d",temp->coeff,temp->power);
   temp=temp->next;
   }
 }

我的代码编译没有显示任何错误或警告，但是我无法打印多项式函数。显示功能仅打印0 x ^ 0或根本不打印任何值，我已经尝试了很多事情，但是没有用，有人可以指出我该怎么做才能更正它。我正在使用C语言。

Answer 1

在这里按值传递指针

create(poly1);

根据您的职能

void create(struct node *head)

然后您设置head

head=newnode;

但是仅修改局部变量head，而您想要修改poly1。

一种可能性是像这样传递双指针：

 create(&poly1);
 ...
 void create(struct node **head)
 ...
 if (*head == NULL)
     *head = newnode;

现在您真正地修改了所需的全局变量poly1。

Answer 2

我的代码编译未显示任何错误或警告

这很奇怪。添加缺少的包含并注释掉getch之后，这是编译器的输出，甚至没有启用（几乎是强制性的）标志-Wall -Wextra：

$ gcc main.c
main.c:4:20: warning: ‘struct node’ declared inside parameter list will not be visible outside of this definition or declaration
    4 | void create(struct node *head);                   //declaring functions
      |                    ^~~~
main.c:5:21: warning: ‘struct node’ declared inside parameter list will not be visible outside of this definition or declaration
    5 | void display(struct node *head);
      |                     ^~~~
main.c: In function ‘main’:
main.c:17:4: warning: implicit declaration of function ‘clrscr’ [-Wimplicit-function-declaration]
   17 |    clrscr();                                              //main function
      |    ^~~~~~
main.c:19:11: warning: passing argument 1 of ‘create’ from incompatible pointer type [-Wincompatible-pointer-types]
   19 |    create(poly1);
      |           ^~~~~
      |           |
      |           struct node *
main.c:4:26: note: expected ‘struct node *’ but argument is of type ‘struct node *’
    4 | void create(struct node *head);                   //declaring functions
      |             ~~~~~~~~~~~~~^~~~
main.c:20:12: warning: passing argument 1 of ‘display’ from incompatible pointer type [-Wincompatible-pointer-types]
   20 |    display(poly1);
      |            ^~~~~
      |            |
      |            struct node *
main.c:5:27: note: expected ‘struct node *’ but argument is of type ‘struct node *’
    5 | void display(struct node *head);
      |              ~~~~~~~~~~~~~^~~~
main.c: At top level:
main.c:24:8: error: conflicting types for ‘create’
   24 |   void create(struct node *head)
      |        ^~~~~~
main.c:4:6: note: previous declaration of ‘create’ was here
    4 | void create(struct node *head);                   //declaring functions
      |      ^~~~~~
main.c:53:7: error: conflicting types for ‘display’
   53 |  void display(struct node *head)
      |       ^~~~~~~
main.c:5:6: note: previous declaration of ‘display’ was here
    5 | void display(struct node *head);

您的编译器似乎已经严重过时。

您的代码似乎有很多问题，但显而易见的是：

newnode=(struct node*)malloc(sizeof(newnode));

强制转换是不好的做法，但不应引起任何实际问题，但是sizeof的论点是错误的。应该是：

newnode=malloc(sizeof(*newnode));