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与联合作为数据类型的双重链接列表的问题

发布于 2025-02-12 05:52:45 字数 3038 浏览 1 评论 0原文

我必须创建与Union作为数据类型的列表，但事实证明，创建我的列表后，只有在我所有列表中重复的相同值。我该如何处理？

你能帮我这个方向吗？

对您来说，美好的一天

结构，联合和功能声明

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Declaration of list
struct list_t{
    struct node_t *head;
    struct node_t *tail;
};
struct node_t {
    union val_t *vals;
    struct node_t *prev;
    struct node_t *next;
};

//Declaration of Union
union val_t{
    int as_int;
    double as_double;
    char *as_str;
};
int insert(struct list_t *list, union val_t*value);

功能定义

//Push to create my list
int push_none(struct list_t *list, union val_t*value){
    if(list==NULL || value==NULL)return 1;
    struct node_t *node = (struct node_t*) malloc(sizeof(struct node_t));
    if(node==NULL)return 2;
    node->vals =value;


    if(list->head==NULL){
        node->next=NULL;
        node->prev=NULL;
        list->head = node;
        list->tail=node;

    } else{
        node->next=NULL;
        node->prev = list->tail;
        list->tail->next = node;
        list->tail = node;
    }
    return 0;
}

main

int main() {
    //initialization of my list

    struct list_t * list = malloc(sizeof(struct list_t));
    list->tail=NULL;
    list->head=NULL;

    printf("Create a double linked list of 3 element :");
    char buf[100],**ptr=NULL;
    int type[3][3];

    union val_t values[4];
    int as_int1;
    double as_double1;
    int i=0;
    //Data input
    while (i<3){

        int j=0;
        while(j<3){
            while(1){
                printf("\n1 - INT\n2 -DOUBLE\n3 - CHAR\nEnter type:");
                scanf("%d", &type[i][j]);
                if(type[i][j]>0 && type[i][j]<4)break;
            }
            printf("\nEnter value :");
            scanf("%s",buf);

            if(type[i][j]==1){
                as_int1 = (int)strtol(buf,ptr,10);
                (values+j)->as_int = as_int1;
            }
            else if(type[i][j]==2){
                as_double1 = strtod(buf,ptr);
                ((values+j)->as_double) = as_double1;
            }
            else if(type[i][j]==3){
                (values + j)->as_str = strdup(buf);
            }
            j++;
        }
        //Creation of the list
        push_node(list,values);
        i++;
    }
    i=0;
    //Display values
    while(list->head!=NULL){
        int j=0;
        while(j<3){
            if(type[i][j]==1){
                printf("%d ",list->head->vals->as_int);
            }
            if(type[i][j]==2){
                printf("%lf ",list->head->vals->as_double);
            }
            if(type[i][j]==3){
                printf("%s ",list->head->vals->as_str);
            }
            j++;
        }
        printf("\n");
        i++;
        list->head=list->head->next;
    }
    return 0;
}
原文

I have to create a list with union as data type but it turns out that after the creation of my list there is only the same value which is repeated on all my list. How Can I handle this?

Can you help me in this direction?

Good day to you

Structure, Union and Function declaration

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Declaration of list
struct list_t{
    struct node_t *head;
    struct node_t *tail;
};
struct node_t {
    union val_t *vals;
    struct node_t *prev;
    struct node_t *next;
};

//Declaration of Union
union val_t{
    int as_int;
    double as_double;
    char *as_str;
};
int insert(struct list_t *list, union val_t*value);

Function definition

//Push to create my list
int push_none(struct list_t *list, union val_t*value){
    if(list==NULL || value==NULL)return 1;
    struct node_t *node = (struct node_t*) malloc(sizeof(struct node_t));
    if(node==NULL)return 2;
    node->vals =value;


    if(list->head==NULL){
        node->next=NULL;
        node->prev=NULL;
        list->head = node;
        list->tail=node;

    } else{
        node->next=NULL;
        node->prev = list->tail;
        list->tail->next = node;
        list->tail = node;
    }
    return 0;
}

Main

int main() {
    //initialization of my list

    struct list_t * list = malloc(sizeof(struct list_t));
    list->tail=NULL;
    list->head=NULL;

    printf("Create a double linked list of 3 element :");
    char buf[100],**ptr=NULL;
    int type[3][3];

    union val_t values[4];
    int as_int1;
    double as_double1;
    int i=0;
    //Data input
    while (i<3){

        int j=0;
        while(j<3){
            while(1){
                printf("\n1 - INT\n2 -DOUBLE\n3 - CHAR\nEnter type:");
                scanf("%d", &type[i][j]);
                if(type[i][j]>0 && type[i][j]<4)break;
            }
            printf("\nEnter value :");
            scanf("%s",buf);

            if(type[i][j]==1){
                as_int1 = (int)strtol(buf,ptr,10);
                (values+j)->as_int = as_int1;
            }
            else if(type[i][j]==2){
                as_double1 = strtod(buf,ptr);
                ((values+j)->as_double) = as_double1;
            }
            else if(type[i][j]==3){
                (values + j)->as_str = strdup(buf);
            }
            j++;
        }
        //Creation of the list
        push_node(list,values);
        i++;
    }
    i=0;
    //Display values
    while(list->head!=NULL){
        int j=0;
        while(j<3){
            if(type[i][j]==1){
                printf("%d ",list->head->vals->as_int);
            }
            if(type[i][j]==2){
                printf("%lf ",list->head->vals->as_double);
            }
            if(type[i][j]==3){
                printf("%s ",list->head->vals->as_str);
            }
            j++;
        }
        printf("\n");
        i++;
        list->head=list->head->next;
    }
    return 0;
}

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评论（1）

岁月苍老的讽刺 2025-02-19 05:52:45

这结合了我在最高评论中所说的话。

  1. 将类型添加到节点结构中
  2. 具有节点结构中值的实例（ not> a指针），
  3. 每个节点只有一个值

是重构代码。它进行了编译并经过了[限制]测试：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Declaration of list
struct list_t {
    struct node_t *head;
    struct node_t *tail;
};

//Declaration of Union
union val_t {
    int as_int;
    double as_double;
    char *as_str;
};

struct node_t {
    int type;
    union val_t vals;
    struct node_t *prev;
    struct node_t *next;
};

int insert(struct list_t *list, union val_t *value);

//Push to create my list
int
push_node(struct list_t *list, union val_t value, int type)
{
    if (list == NULL)
        return 1;

    struct node_t *node = malloc(sizeof(*node));
    if (node == NULL)
        return 2;

    node->vals = value;
    node->type = type;

    if (list->head == NULL) {
        node->next = NULL;
        node->prev = NULL;
        list->head = node;
        list->tail = node;

    }
    else {
        node->next = NULL;
        node->prev = list->tail;
        list->tail->next = node;
        list->tail = node;
    }

    return 0;
}

int
main(void)
{

    // initialization of my list
    struct list_t *list = malloc(sizeof(*list));
    list->tail = NULL;
    list->head = NULL;

    printf("Create a double linked list of 3 element :");
    char buf[100], **ptr = NULL;
    int type;

    union val_t values;
    int i = 0;

    // Data input
    for (;  i < 3;  ++i) {
        while (1) {
            printf("\n1 - INT\n2 -DOUBLE\n3 - CHAR\nEnter type:");
            scanf("%d", &type);
            if (type > 0 && type < 4)
                break;
        }

        printf("\nEnter value :");
        scanf("%s", buf);

        switch (type) {
        case 1:
            values.as_int = strtol(buf, ptr, 10);
            break;

        case 2:
            values.as_double = strtod(buf, ptr);
            break;

        case 3:
            values.as_str = strdup(buf);
            break;
        }

        // Creation of the list
        push_node(list, values, type);
    }

    // Display values
    for (struct node_t *node = list->head;  node != NULL;  node = node->next) {
        switch (node->type) {
        case 1:
            printf("%d ", node->vals.as_int);
            break;
        case 2:
            printf("%f ", node->vals.as_double);
            break;
        case 3:
            printf("%s ", node->vals.as_str);
            break;
        }
        printf("\n");
    }

    return 0;
}

这是一些测试输出：

Create a double linked list of 3 element :
1 - INT
2 -DOUBLE
3 - CHAR
Enter type:1

Enter value :23

1 - INT
2 -DOUBLE
3 - CHAR
Enter type:2

Enter value :37.889

1 - INT
2 -DOUBLE
3 - CHAR
Enter type:3

Enter value :jwelrjlewjrowejrowier
23 
37.889000 
jwelrjlewjrowejrowier

这是一个更清洁的版本。特别是，node_push已经简化了：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Declaration of list
struct list_t {
    struct node_t *head;
    struct node_t *tail;
};

//Declaration of Union
union val_t {
    int as_int;
    double as_double;
    char *as_str;
};

struct node_t {
    int type;
    union val_t vals;
    struct node_t *prev;
    struct node_t *next;
};

int insert(struct list_t *list, union val_t *value);

//Push to create my list
int
push_node(struct list_t *list, union val_t value, int type)
{
    if (list == NULL)
        return 1;

    struct node_t *node = malloc(sizeof(*node));
    if (node == NULL)
        return 2;

    node->vals = value;
    node->type = type;

    node->next = NULL;
    node->prev = list->tail;

    if (list->head == NULL)
        list->head = node;
    else
        list->tail->next = node;

    list->tail = node;

    return 0;
}

int
main(void)
{

    // initialization of my list
    struct list_t *list = calloc(1,sizeof(*list));

    printf("Create a double linked list of 3 element :");
    char buf[100], **ptr = NULL;
    int type;

    union val_t values;
    int i = 0;

    // Data input
    for (;  i < 3;  ++i) {
        while (1) {
            printf("\n1 - INT\n2 -DOUBLE\n3 - CHAR\nEnter type:");
            scanf("%d", &type);
            if (type > 0 && type < 4)
                break;
        }

        printf("\nEnter value :");
        scanf("%s", buf);

        switch (type) {
        case 1:
            values.as_int = strtol(buf, ptr, 10);
            break;

        case 2:
            values.as_double = strtod(buf, ptr);
            break;

        case 3:
            values.as_str = strdup(buf);
            break;
        }

        // Creation of the list
        push_node(list, values, type);
    }

    // Display values
    printf("\n");
    printf("List:\n");
    for (struct node_t *node = list->head;  node != NULL;  node = node->next) {
        printf("type: %d ",node->type);

        switch (node->type) {
        case 1:
            printf("%d ", node->vals.as_int);
            break;
        case 2:
            printf("%f ", node->vals.as_double);
            break;
        case 3:
            printf("%s ", node->vals.as_str);
            break;
        }

        printf("\n");
    }

    return 0;
}

尽管有些不同，但您可能希望在C：写作a C 中的“通用”结构印刷方法

This incorporates what I was saying in my top comments.

  1. Add type to node struct
  2. Have an instance of the value in the node struct (not a pointer)
  3. Only one value per node

Here is the refactored code. It compiles and has been [cursorily] tested:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Declaration of list
struct list_t {
    struct node_t *head;
    struct node_t *tail;
};

//Declaration of Union
union val_t {
    int as_int;
    double as_double;
    char *as_str;
};

struct node_t {
    int type;
    union val_t vals;
    struct node_t *prev;
    struct node_t *next;
};

int insert(struct list_t *list, union val_t *value);

//Push to create my list
int
push_node(struct list_t *list, union val_t value, int type)
{
    if (list == NULL)
        return 1;

    struct node_t *node = malloc(sizeof(*node));
    if (node == NULL)
        return 2;

    node->vals = value;
    node->type = type;

    if (list->head == NULL) {
        node->next = NULL;
        node->prev = NULL;
        list->head = node;
        list->tail = node;

    }
    else {
        node->next = NULL;
        node->prev = list->tail;
        list->tail->next = node;
        list->tail = node;
    }

    return 0;
}

int
main(void)
{

    // initialization of my list
    struct list_t *list = malloc(sizeof(*list));
    list->tail = NULL;
    list->head = NULL;

    printf("Create a double linked list of 3 element :");
    char buf[100], **ptr = NULL;
    int type;

    union val_t values;
    int i = 0;

    // Data input
    for (;  i < 3;  ++i) {
        while (1) {
            printf("\n1 - INT\n2 -DOUBLE\n3 - CHAR\nEnter type:");
            scanf("%d", &type);
            if (type > 0 && type < 4)
                break;
        }

        printf("\nEnter value :");
        scanf("%s", buf);

        switch (type) {
        case 1:
            values.as_int = strtol(buf, ptr, 10);
            break;

        case 2:
            values.as_double = strtod(buf, ptr);
            break;

        case 3:
            values.as_str = strdup(buf);
            break;
        }

        // Creation of the list
        push_node(list, values, type);
    }

    // Display values
    for (struct node_t *node = list->head;  node != NULL;  node = node->next) {
        switch (node->type) {
        case 1:
            printf("%d ", node->vals.as_int);
            break;
        case 2:
            printf("%f ", node->vals.as_double);
            break;
        case 3:
            printf("%s ", node->vals.as_str);
            break;
        }
        printf("\n");
    }

    return 0;
}

Here is some test output:

Create a double linked list of 3 element :
1 - INT
2 -DOUBLE
3 - CHAR
Enter type:1

Enter value :23

1 - INT
2 -DOUBLE
3 - CHAR
Enter type:2

Enter value :37.889

1 - INT
2 -DOUBLE
3 - CHAR
Enter type:3

Enter value :jwelrjlewjrowejrowier
23 
37.889000 
jwelrjlewjrowejrowier

Here is a slightly more cleaned up version. In particular, node_push has been simplified:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

//Declaration of list
struct list_t {
    struct node_t *head;
    struct node_t *tail;
};

//Declaration of Union
union val_t {
    int as_int;
    double as_double;
    char *as_str;
};

struct node_t {
    int type;
    union val_t vals;
    struct node_t *prev;
    struct node_t *next;
};

int insert(struct list_t *list, union val_t *value);

//Push to create my list
int
push_node(struct list_t *list, union val_t value, int type)
{
    if (list == NULL)
        return 1;

    struct node_t *node = malloc(sizeof(*node));
    if (node == NULL)
        return 2;

    node->vals = value;
    node->type = type;

    node->next = NULL;
    node->prev = list->tail;

    if (list->head == NULL)
        list->head = node;
    else
        list->tail->next = node;

    list->tail = node;

    return 0;
}

int
main(void)
{

    // initialization of my list
    struct list_t *list = calloc(1,sizeof(*list));

    printf("Create a double linked list of 3 element :");
    char buf[100], **ptr = NULL;
    int type;

    union val_t values;
    int i = 0;

    // Data input
    for (;  i < 3;  ++i) {
        while (1) {
            printf("\n1 - INT\n2 -DOUBLE\n3 - CHAR\nEnter type:");
            scanf("%d", &type);
            if (type > 0 && type < 4)
                break;
        }

        printf("\nEnter value :");
        scanf("%s", buf);

        switch (type) {
        case 1:
            values.as_int = strtol(buf, ptr, 10);
            break;

        case 2:
            values.as_double = strtod(buf, ptr);
            break;

        case 3:
            values.as_str = strdup(buf);
            break;
        }

        // Creation of the list
        push_node(list, values, type);
    }

    // Display values
    printf("\n");
    printf("List:\n");
    for (struct node_t *node = list->head;  node != NULL;  node = node->next) {
        printf("type: %d ",node->type);

        switch (node->type) {
        case 1:
            printf("%d ", node->vals.as_int);
            break;
        case 2:
            printf("%f ", node->vals.as_double);
            break;
        case 3:
            printf("%s ", node->vals.as_str);
            break;
        }

        printf("\n");
    }

    return 0;
}

Although somewhat different, you may want to have a look at my answer on "generic" structs in C: Writing a 'generic' struct-print method in C

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