程序在printPoly（）函数中遇到无限循环

Question

这个程序应该从用户那里读取2个多项式，然后打印出求和；一旦两个多项式都被输入，就会陷入无限循环。

>

为了存储一个多项式，使用了一个带有指向下一项（节点）的*plink前向声明的结构

每个多项式都有一个头节点，它有助于以循环方式遍历列表。

因为头节点不包含任何与多项式“相关”的数据，所以它们的指数字段被初始化为-1

#include"stdio.h"
#include"stdlib.h"

#define SIZE 10

#define TRUE 1
#define FALSE 0
struct poly
{
int exp;
float coef;
struct poly *plink;
};

typedef struct poly PNODE;

PNODE* getNode()
{
PNODE *ptr= (PNODE*)malloc(sizeof(PNODE));
if(ptr==NULL)
{
    printf("\nNo Memory\n");
    return;
}
return ptr;
}
    void readPoly(PNODE *phead, int n)
    {
        int i, ex;
        float co;
        PNODE *temp,*newnode;
        temp=phead;
        for(i=0;i<n;i++)
            {
            printf("\nEnter the EXPONENT and COEFFICIENT for term %d\n",i);
            scanf("%d%f",&ex,&co);

            newnode=getNode();

            newnode->exp=ex;
            newnode->coef=co;

            temp->plink=newnode;
            temp=newnode;
        }
     temp->plink=phead;
     return;
    }

    void printPoly(PNODE* phead)
    {
        PNODE *i=phead;
        for(i=phead->plink;i->plink!=phead;i=phead->plink)
            {
                printf("%2f(x^%d)+",i->coef,i->exp);
            }
            printf("\n");
    }

int compare(int a, int b)
    {
        if(a>b)
        {
           return 1;
        }
         if(b>a)
        {
         return -1;
        }
       if(a==b)
        {
            return 0;
        }
    }

void attach(int exp, float coef, PNODE **add)
{
    PNODE *temp;
    temp=getNode();

temp->exp=exp;
temp->coef=coef;

(*add)->plink=temp;
*add=temp;
    return;
}

PNODE* addPoly(PNODE *phead1, PNODE *phead2)
{
PNODE *startA, *res, *tempres;
int done= FALSE;
float sum= 0.0;

startA=phead1;

phead1=phead1->plink;
phead2=phead2->plink;

res=getNode();
res->exp=-1;
tempres=res;

do
{
    switch(compare(phead1->exp,phead2->exp))
    {
         case 1: attach(phead1->coef,phead1->exp,&tempres);
                phead1=phead1->plink;
                break;
         case -1: attach(phead2->coef,phead2->exp,&tempres);
                phead2=phead2->plink;
                break;
         case 0: if(startA==phead1)
                        {
                            done=TRUE;
                        }
                else
                {
                    sum=phead1->coef+phead2->coef;
                    if(sum)
                    {
                        attach(sum,phead1->coef,&tempres);
                    }
                   phead1=phead1->plink;
                   phead2=phead2->plink;
                }
                break;

    }
}while(!done);
tempres->plink=res;
return res;
}
int main()
{
int n1, n2;
PNODE *phead1, *phead2,*result;

phead1=(PNODE*)malloc(sizeof(PNODE));
phead2=(PNODE*)malloc(sizeof(PNODE));

phead1->exp=-1;
phead2->exp=-1;

phead1->plink=phead1;
phead2->plink=phead2;

printf("\nEnter the number of terms in Polynomial 1\n");
scanf("%d",&n1);
readPoly(phead1,n1);
printf("\nEnter the number of terms in Polynomial 2\n");
scanf("%d",&n2);
readPoly(phead2,n2);

printf("The first Polynomial is \n");
printPoly(phead1);
printf("The second Polynomial is \n");
printPoly(phead2);

printf("Result is\n");
result=addPoly(phead1,phead2);
printPoly(result);

return 0;
}

Answer 1

for(i=phead->plink;i->plink!=phead;i=phead->plink)

i总是被赋予相同的值，因此循环永远不会结束。应该是i=i->plink移动到下一个节点。