使用循环增加char值到链表[C] [英] Adding Char Value to a Linked List using Looping [C]

问题描述

我有一个即时通讯目前正试图将值添加到它的链接列表。但是，我必须设置不正确我的指针或者有一些与内存分配回事。

我要令牌添加到列表中，但每次有一个新的循环中的数据重叠。例如：

第1次：

REPL>一
一个

第2次：

REPL>乙
b

乙

注意如何公正消失了，我想保持previous值，而在新的价值增加。

下面是我的code迄今：

 结构节点{
    INT VAL;
    结构节点*接下来的;
};结构节点*头= NULL;
结构节点* CUR = NULL;结构节点* create_list（int值）
{
    结构节点* PTR =（结构节点*）malloc的（的sizeof（结构节点））;
    如果（NULL == PTR）返回NULL;
    ptr-＆GT; VAL =价值;
    ptr-＆gt;接着= NULL;    ptr-＆gt;接下来=头;
    头= PTR;    返回PTR;
};结构节点*插入（int值）
{
    如果（NULL ==头）
        回报（create_list（值））;    结构节点* PTR =（结构节点*）malloc的（的sizeof（结构节点））;
    ptr-＆GT; VAL =价值;
    ptr-＆gt;接着= NULL;    ptr-＆gt;接下来=头;
    头= PTR;    返回PTR;};无效print_list（无效）
{
    结构节点* PTR =头;    而（PTR！= NULL）{
        的printf（％S \\ n，ptr-＆GT; VAL）;
        PTR = ptr-＆gt;接下来，
    }    返回;
}结构EXP {
    int型的;
    工会{
        INT NUM;
        字符名称;
        双小数;
        焦炭strq;
    }值;
};
INT主（INT ARGC，CHAR *的argv []）
{
    而（1）{
        输出（REPL＆GT;）;
        字符*存储[30];
        字符* TOK;
        焦克;
        炭缓冲液[20];
        INT POS = 0，I;
        与fgets（缓冲区，缓冲区尺寸，标准输入）;        TOK = strtok的（缓冲，）;        而（TOK）{
            POS ++;
            存储[POS] = TOK;
            create_list（存储[POS]）;
            TOK =的strtok（NULL，）;
        }        print_list（）;
    }
}
 

解决方案

我看到你的code以下问题：

1. 在 print_list ，您可能需要更改的printf（％S \\ n，ptr-＆GT; VAL）; 到的printf（％C \\ N，ptr-＆GT; VAL）; 如果你想在节点作为字符打印值


2. 我不知道你为什么使用它之前递增 POS 。你可能意味着该行 create_list（存储[POS]）后，增加它;


3. 参数类型为 create_list 是 INT 。你是通过一个的char * 它。也许你的意思是通过存储[POS] [0] 。


4. 您可能也意味着 TOK = strtok的（TOK，）; 。否则，，而循环没有做任何好处。

在我做我的电脑这些更改到code，程序的表现就像你所期望它O操作。

I have a linked list which Im currently trying to add values to it. But I must have set my pointers incorrectly or there is something going on with the memory allocation.

I want to add tokens to the list but everytime there is a new loop the data overlaps. For example:

1st time:

repl> a a

2nd time:

repl> b b

b

Notice how the a just disappears, I want to keep the previous values while adding in new values.

Here's my code so far:

struct node {
    int val;
    struct node *next;
};

struct node *head = NULL;
struct node *cur = NULL;

struct node* create_list (int value)
{
    struct node *ptr = (struct node*) malloc(sizeof (struct node));
    if (NULL == ptr) return NULL;
    ptr->val = value;
    ptr->next = NULL;

    ptr->next = head;
    head = ptr;

    return ptr;
};

struct node* insertion (int value)
{
    if (NULL == head)
        return (create_list(value));

    struct node *ptr = (struct node*)malloc(sizeof(struct node));
    ptr->val = value;
    ptr->next = NULL;

    ptr->next = head;
    head = ptr;

    return ptr;

};

void print_list(void)
{
    struct node *ptr = head;

    while(ptr != NULL) {
        printf(" %s\n",ptr->val);
        ptr = ptr->next;
    }

    return;
}

struct exp {
    int type;
    union {
        int num;
        char name;
        double decimal;
        char strq;
    } value;
};


int main(int argc, char *argv[])
{
    while(1) {
        printf("repl>");
        char *storage [30];
        char* tok;
        char g;
        char buffer[20];
        int pos = 0, i;
        fgets(buffer,sizeof(buffer),stdin);

        tok = strtok(buffer," ");

        while(tok) {
            pos++;
            storage[pos] = tok;
            create_list(storage[pos]);
            tok = strtok(NULL," ");
        }

        print_list();
    }
}

解决方案

I see the following problems in your code:

1. In print_list, you may want to change printf(" %s\n",ptr->val); to printf(" %c\n",ptr->val); if you want to print the value at the node as a character.
2. I don't know why you are incrementing pos before using it. You probably meant to increment it after the line create_list(storage[pos]);.
3. The argument type to create_list is an int. You are passing a char * to it. Perhaps you meant to pass storage[pos][0].
4. You probably also meant tok = strtok(tok, " ");. Otherwise, the while loop is not doing you any good.

After I made those changes to your code in my computer, the program behaved like you expected it o.

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