关于函数内参数地址的改变或返回

Question

题目
【问题描述】

编写一个程序实现两个一元多项式相乘。

【输入形式】

首先输入第一个多项式中系数不为0的项的系数和指数，以一个空格分隔。且该多项式中各项的系数均为0或正整数，系数和最高幂次不会超过int类型的表示范围。对于多项式的输入方法如下：
an n a n-1 n-1 … a1 1 a0 0
即相邻两个整数分别表示表达式中一项的系数和指数。在输入中只出现系数不为0的项。最后一项的指数后没有空格，只有一个回车换行符。
按照上述方式再输入第二个多项式。

【输出形式】

将运算结果输出到屏幕。将系数不为0的项按指数从高到低的顺序输出，每次输出其系数和指数，均以一个空格分隔，最后一项的指数后也可以有一个空格。

【样例输入】

10 80000 2 6000 7 300 5 10 18 0
3 6000 5 20 8 10 6 0

【样例输出】

30 86000 50 80020 80 80010 60 80000 6 12000 21 6300 10 6020 31 6010 66 6000 35 320 56 310 42 300 25 30 130 20 174 10 108 0

提示：利用链表存储多项式的系数和指数。

问题
笔者（C语言初学者）根据现有知识通过结构体编程，但发现其中部分模块出现问题，根据调试观察，问题出在函数内的参数地址问题，笔者不清楚函数参数如何改变并返回。下面附上代码，其中函数cal是问题所在。

#include<stdio.h>
#include<malloc.h>

typedef struct list
{
    int coe;
    int sub;
    struct list* next;
}num;

void cal(num* head,num* turn, num* end);

void order(num* head, num* end);

void print(num* head, num* end);

int main()
{
    int ch;
    int coe, sub;
    num* p = NULL;
    num* turn = NULL;
    num* node = NULL;
    num* head = NULL;
    num* end = NULL;

    do
    {
        scanf_s("%d %d", &coe, &sub);
        p = (num*)malloc(sizeof(num*));
        if (head == NULL)
            head = p;
        else
        {
            node->next = p;
        }
        if(p)
        {
            p->coe = coe;
            p->sub = sub;
        }
        node = p;
    } while ((ch = getchar()) != '\n');
    turn = node;
    do
    {
        scanf_s("%d %d", &coe, &sub);
        p = (num*)malloc(sizeof(num*));
        if(p&&node)
        {
            node->next = p;
            p->coe = coe;
            p->sub = sub;
        }
        
        node = p;
    } while ((ch = getchar()) != '\n');

    end = node;
    cal(head,turn,end);
    order(head,end);
    print(head,end);

    return 0;
}

void cal(num* head,num* turn, num* end)
{
    num* p = NULL;
    num* p1 = head;
    num* p2 = turn;
    int i=0;
    int j = 0;
    do
    {
        p2 = turn;
        if (i==1)
        {
            p1 = p1->next;
        }
        i = 1;
        do
        {
            p2 = p2->next;
            p = (num*)malloc(sizeof(num*));
            if(j==0)
            {            
                head = p;
                j = 1;
            }
            
            if(p)
            {
                p->coe = p1->coe * p2->coe;
                p->sub = p1->sub * p2->sub;
            }
            
        }while (p2 != end);
        
    } while (p1 != turn);
    end = p;
}

void order(num* head, num* end)
{
    num* p1 = head;
    num* p2 = head;
    num* p3 = head;

    int i;
    int j=0;
    int q= 0;

    

    do
    {
        if (j == 1)
        {
            p1 = p1->next;
            p2 = p1;
            p3 = p1;
        }
        j = 1;
        q = 0;
        
        do
        {
            
            if (q == 1)
            {
                p3 = p3->next;
            }
            p2 = p3->next;
            q = 1;
            if (p1->sub < p2->sub)
            {
                i = p1->sub;
                p1->sub = p2->sub;
                p2->sub = i;
            }
            if (p1->sub == p2->sub)
            {
                p1->coe += p2->coe;
                if(p2!=end)
                {
                    p3->next = p2->next;
                }
                else
                {
                    end = p3;
                }
            }
        } while (p2 != end);
    } while (p1 != end);
}

void print(num* head, num* end)
{
    num* p = head;
    int i = 0;

    do
    {
        if (i == 1)
        {
            p = p->next;
        }
        i = 1;
        printf("%d %d ", p->coe, p->sub);
    } while (p != end);
}

Answer 1

函数内分配的地址要通过双指针参数或者返回值传出函数

#include <stdio.h>
#include <malloc.h>

typedef struct list
{
   int coe;
   int sub;
   struct list *next;
} num;

num *cal(num *heada, num *headb, num **headc);

void order(num *head);

void print(num *p);

int main()
{
   int ch;
   int coe, sub;
   num *p, *node;
   num *heada = NULL;
   num *headb = NULL;
   num *headc = NULL;

   do
   {
      scanf_s("%d%d", &coe, &sub);
      p = (num *)malloc(sizeof(num)); //
      p->next = NULL;
      if (p)
      {
         p->coe = coe;
         p->sub = sub;
      }

      if (heada == NULL)
         heada = p;
      else
         node->next = p;

      node = p;
   } while ((ch = getchar()) != '\n');

   do
   {
      scanf_s("%d%d", &coe, &sub);
      p = (num *)malloc(sizeof(num)); //
      p->next = NULL;
      if (p)
      {
         p->coe = coe;
         p->sub = sub;
      }

      if (headb == NULL)
         headb = p;
      else
         node->next = p;

      node = p;
   } while ((ch = getchar()) != '\n');

   cal(heada, headb, &headc);
   order(headc);
   print(headc);

   return 0;
}

num *cal(num *heada, num *headb, num **headc)
{
   num *p = NULL;
   num *p1 = heada;
   num *p2 = headb;
   num *p3 = NULL;

   while (p1 != NULL)
   {
      p2 = headb;
      while (p2 != NULL)
      {
         p = (num *)malloc(sizeof(num)); //
         p->next = NULL;
         if (p)
         {
            p->coe = p1->coe * p2->coe;
            p->sub = p1->sub + p2->sub; //
         }
         if (p3 == NULL)
            *headc = p3 = p;

         else
            p3->next = p;

         p3 = p;

         p2 = p2->next;
      }
      p1 = p1->next;
   }

   return *headc;
}

void order(num *head)
{
   num *p1 = head;
   num *p2 = head->next;
   num *p3 = p1;
   int t;
   while (p1 != NULL)
   {
      p2 = p1->next;
      p3 = p1;
      while (p2 != NULL)
      {
         if (p1->sub < p2->sub)
         {
            t = p1->sub;
            p1->sub = p2->sub;
            p2->sub = t;
            t = p1->coe;
            p1->coe = p2->coe;
            p2->coe = t;
         }
         if (p1->sub == p2->sub)
         {
            p1->coe += p2->coe;
            p3->next = p2->next;
            free(p2);
            p2 = p3->next;
            continue;
         }
         p3 = p2;
         p2 = p2->next;
      }
      p1 = p1->next;
   }
}

void print(num *p)
{
   while (p != NULL)
   {
      printf("%d %d ", p->coe, p->sub);
      p = p->next;
   }
}