C - 将数组传递给链表的节点

Question

我有一个已填充整数的二维数组，准备切割成行并进行处理，我需要将每一行(一维数组)传递给链表的一个节点。每个节点看起来像这样:

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

以这种方式添加和链接新节点:

struct node *addnode(int *val, struct node *next, int columns)
{
    struct node *tnode;
    tnode = (struct node*)malloc(sizeof(*tnode));
    if(tnode != NULL) {
        tnode->val = malloc(sizeof(int) * columns);
        memcpy(tnode->val, val, sizeof(int) * columns);
        tnode->val = val;
        tnode->next = next;
    };
    return tnode;
}

将填充每个节点的程序片段大致如下所示:

int table[rows][columns], i, j;
for (i = 0; i < rows; i++){
    head = addnode(*table, head, columns);
    for (j = 0; j < columns; j++){
        scanf("%d",&table[i][j]);
        head->val[j] = table[j];
        printf("%d ",head->val[j]);
    };
    puts("\n");
};  

我不确定如何在指示的地方进行:

1. 那是整个节点的 malloc，但是我应该如何处理 val 的 malloc？我知道每个节点中应该有的表的长度，它是 columns，在 main 函数中获取。我应该在哪里为其分配内存？
2. 此行上方是我为单行整数分配足够(列)内存的地方。这是一个不错的选择吗？
3. 这个，在前面的循环中应该用足够的 i 行二维数组 table 填充当前的 head->val[j]，但它看起来太好了是真实的。我可以就这样离开吗？

编辑:我在某些地方更正了它，但在尝试对其进行排序后，它返回垃圾。我将在这里转储大部分代码:

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

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

struct node *addnode(int *val, struct node *next, int columns);
struct node *mergesort(struct node *head, int column);
struct node *merge(struct node *head_one, struct node *head_two, int column);

int main(int argc, char *argv[])
{

    struct node *head;
    struct node *current;
    struct node *next;
    int symbol = 0;
    int columns = 0;
    //int columns = atoi(argv[1]); //until sorting works, I'll keep it at 0
    int rows = 0;
    head = NULL;
    int column = 0; //temporary until I find the way to send one argument during executing it under linux like so 'name_of_program columns < test.txt'
    int lastSpace = 0;

    do {
        symbol = fgetc(stdin);

        if (rows == 0 && (lastSpace == 0 && (isspace(symbol) || feof(stdin)))) {
            columns++;
            lastSpace = 1;
        } else if (!isspace(symbol)) {
            lastSpace = 0;
        }
        if (symbol == '\n' || feof(stdin)) {
            rows++;
        };      
    } while (symbol != EOF);

    if (ferror(stdin))
    {
        printf("Error on reading from file.\n");
    } else {
        printf("The file contains %d row(s) and %d column(s).\n", rows, columns);
    };

    rewind(stdin); //I have heard conflicting opinions on that, but in this case it works, and in the end it's a school project, not commercial code

    int table[rows][columns], i, j;
    for (i = 0; i < rows; i++){
        head = addnode(*table, head, columns);
        for (j = 0; j < columns; j++){
            scanf("%d",&table[i][j]);
            head->val[j] = table[i][j];
            printf("%d ",head->val[j]);
        };
        puts("\n");
    };  


    head = mergesort(head, column);

    for(current = head; current != NULL; current = current->next){
        for (j = 0; j < columns; j++){
            printf("%d ", current->val[j]);
        };
        puts("\n");
    };

    for(current = head; current != NULL; current = next)
      next = current->next, free(current);
    return 0;
};


struct node *addnode(int *val, struct node *next, int columns)
{
    struct node *tnode;
    tnode = (struct node*)malloc(sizeof(*tnode));
    if(tnode != NULL) {
        tnode->val = malloc(sizeof(int) * columns);
        memcpy(tnode->val, val, sizeof(int) * columns);
        tnode->val = val;
        tnode->next = next;
    };
    return tnode;
}

struct node *mergesort(struct node *head, int column)
{
    struct node *head_one;
    struct node *head_two;
    if((head == NULL) || (head->next == NULL))
        return head;
    head_one = head;
    head_two = head->next;
    while((head_two != NULL) && (head_two->next != NULL)) {
        head = head->next;
        head_two = head->next->next;
    };
    head_two = head->next;
    head->next = NULL;
    return merge(mergesort(head_one, column), mergesort(head_two, column), column);
}

struct node *merge(struct node *head_one, struct node *head_two, int column)
{
    struct node *head_combined;
    if(head_one == NULL)
        return head_two;
    if(head_two == NULL)
        return head_one;
    if(head_one->val[column] < head_two->val[column]) {
        head_combined = head_one;
        head_combined->next = merge(head_one->next, head_two, column);
    } else {
        head_combined = head_two;
        head_combined->next = merge(head_one, head_two->next, column);
    };
    return head_combined;
}

我在 Unix 中运行它是这样的:

name_of_program < test.txt

test.txt 的结构为 http://pastebin.com/WL5brutf

Answer 1

1) 您正在将 int *val 传递给函数，并在节点中使用它本身。如果你想丢弃传递给函数的 val 并想保留它的另一个副本，你需要 malloc 内存。正如您所说，您知道数组中有多少个元素，因此您可以分配那么多元素，只需将内存从 val 复制为

 tnode->val = malloc(sizeof(int) * num_of_elements); //replace num_of_elements with appropriate variable/constant
    memcpy(tnode->val, val, sizeof(int) * num_of_elements);

2) 是的，这是正确的地方。

3) 是的，您可以用这种方式为 head 指向的当前节点赋值。您可能希望在 j for 循环结束后移动到下一个节点并在其 val 中分配新值。