Question

我不确定我是否听起来很笨，但仍然......我想使用链接列表开发一个程序，我可以添加5个多项式。我能提出的代码是......

struct node
{
    int exp,coeff;
    struct node *next;
} *start[5]={NULL};

void read()
{
    struct node *current,*newnode;
    int n,i,max,j;
    printf("How many poly?");
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        printf("\nDetails of poly %d\n",i);
        printf("Max degree:");
        scanf("%d",&max);
        for(j=max;j>=0;j--)
        {
            newnode=(struct node *)malloc(sizeof(newnode));
            printf("\nEnter coeff of %d:",j);
            scanf("%d",&newnode->coeff);
            newnode->exp=j;
            if(start[i]==NULL)
            {
                printf("a%d",start[i]);
                start[i]=newnode;
                newnode->next=NULL;
                printf("%d,%d",start[i],start[i]->next);
            }
            else if(start[i]->next==NULL)
            {
                printf("b");
                start[i]->next=newnode;
                newnode->next=NULL;
                printf("%d",start[i]);
            }
            else
            {
                printf("c");
                current=start[i];
                while(current->next!=NULL)
                    current=current->next;
                current->next=newnode;
                newnode->next=NULL;
                printf("%d",start[i]);
            }   
        }
    }   
}

void sum()
{
    struct node *curr1,*curr2,*tmp;
    int i,co,ex;
    curr1=start[0];
    while(curr1!=NULL)
    {   
        for(i=1;i<5;i++)
        {
            co=0;
            if(curr1==start[i])
                tmp=start[i];
            else
                tmp=tmp->next;
            curr2=tmp;
            if(curr1->exp==curr2->exp)
            {
                if(co==0)
                    co=co+curr1->coeff+curr2->coeff;
                else
                    co=co+curr2->coeff;
                ex=curr1->exp;
                break;
            }
            else if(curr1->exp>curr2->exp)
            {   
                ex=curr1->exp;
                if(co==0)
                    co=co+curr1->coeff;
                else
                    break;
            }
            else
            {   
                ex=curr2->exp;
                if(co==0)
                    co=co+curr2->coeff;
                else
                    break;
            }
        }
        printf("%dX^%d",co,ex);
        curr1=curr1->next;
    }
}

int main()
{
    read();
    printf("result:\n");
    sum();
    getch();
    return 0;
}

但是在执行时，只有第一个多项式得到read.ie，'i'循环只工作一次。我试着检查我的阅读功能，注意第二个条件不起作用。这个，这个：

else if(start[i]->next==NULL)    
{
    printf("b");
    start[i]->next=newnode;
    newnode->next=NULL;
    printf("%d",start[i]);
}

打印b是检查程序是否进入此块LOL。有人帮我这个... ...

Answer 1

这部分

newnode=(struct node *)malloc(sizeof(newnode));  // This will allocate space for 
                                                 // a pointer to struct node

应该是

newnode=(struct node *)malloc(sizeof(struct node));  // This will allocate space for 
                                                     // a struct node

或更好，如@chux

所示

newnode = malloc(sizeof *newnode);         // This will allocate space for 
                                           // whatever newnode points to. In
                                           // this case a struct node

不确定它会有所作为，但我会改变：

start[5]={NULL};

到

start[5]={NULL, NULL, NULL, NULL, NULL};

N多项式加法不起作用

1 个答案: