Question

我昨天开始使用C编程并试图实现一个链表（实际上非常基本）。

到目前为止，除了释放列表外，一切都很好。

首先，这是我的代码：

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

/*
 * Struct for elements of the linked list
 */
struct element {
    struct element* nextelement;
    int value;
};

/*
 * Struct for a list itself
 */
struct liste {
    struct element* headelement;
};

/*
 * Function to create new elements
 */
struct element* createelement(int value) {
    struct element* dummyelement = malloc(sizeof(struct element));
    dummyelement->nextelement = NULL;
    dummyelement->value = value;
    return dummyelement;
}

/*
 * Function to create new (empty) lists
 */
struct liste* createlist() {
    struct liste* dummylist = malloc(sizeof(struct liste));
    dummylist->headelement = NULL;
    return dummylist;
}

/*
 * Add an element to a given list
 */
void addelement(struct liste* liste, int value) {

    struct element* dummyelement = createelement(value);
    if (liste->headelement == NULL) {
        liste->headelement = dummyelement;
    } else {
        struct element* iterelement = liste->headelement;
        while (iterelement->nextelement != NULL) {
            iterelement = iterelement->nextelement;
        }
        iterelement->nextelement = dummyelement;
    }

}

/*
 * Plot the elements of a given list
 */
void plotlist(struct liste* liste) {
    if (liste->headelement != NULL) {
        struct element* iterelement = liste->headelement;
        printf("%d\n", iterelement->value);
        while (iterelement->nextelement != NULL) {
            iterelement = iterelement->nextelement;
            printf("%d\n", iterelement->value);
        }
    } else {
        printf("Where is my head?\n");
    }
}

/*
 * This should completely remove the list, but it fails...
 */
void removelist(struct liste* liste) {
    if (liste->headelement != NULL) {
        struct element* iterelement = liste->headelement;
        struct element* nextelement = iterelement->nextelement;
        free(iterelement);
        while (nextelement != NULL) {
            iterelement = nextelement;
            nextelement = iterelement->nextelement;
            free(iterelement);
        }
    }

    free(liste);
}

int main(void) {

    /*
     * Creates a new list
     * Plots the (empty) list
     * Adds two elements to the list
     * Plots the list again
     * Removes the list
     * Last plot shouldn't really happen, but it does.
     */
    struct liste* mylist = createlist();
    printf("First plot.\n");
    plotlist(mylist);
    addelement(mylist, 1);
    addelement(mylist, 2);
    printf("Second plot.\n");
    plotlist(mylist);
    removelist(mylist);
    printf("Third plot.\n");
    plotlist(mylist);

    return 0;

}

我得到以下输出：

First plot.
Where is my head?
Second plot.
1
2
Third plot.
33

很明显，＆＃33; 33＆＃39;是我的问题。我真的不知道这可能是什么输出...此外，我不知道为什么我的重新组合＆＃39;没有正常工作。我释放了我用“malloc”分配的所有东西。我做错了什么？

Answer 1

您不在plotlist内查看您的列表是否确实存在。您直接尝试访问headelement。更好地做到以下几点：

void plotlist(struct liste* liste) {
    if(liste == NULL){
        printf("This list does not even exist.\n");
    }
    else if (liste->headelement != NULL) {
        // ...
    } else {
        printf("Where is my head?\n");
    }
}

由于removelist也会释放列表结构，因此最好将指针设置为main中的NULL。否则，在访问释放的内存时会出现未定义的行为。

int main(void) {
    // ...
    removelist(mylist);
    mylist = NULL;
    printf("Third plot.\n");
    plotlist(mylist);

    return 0;

}

试图“免费”＃39; C中的链表

1 个答案: