在循环中，链表似乎是可以的，但在循环之外是空的

线条

if (first == NULL) { 
    first = newCar; // the new node is the only node so it is the first node
    last = newCar; // the new node is the only node so it is also the last node
} else {
    last->next = newCar; // the new node is next node of the previous last node
    last = newCar; // the new node is now the last node
}

在函数中仅修改变量，在函数中修改变量。它们不会修改变量和函数。readCarsfirstlastreadCarsfirstlastmain

因此，函数中的变量将保留，因此，就该函数而言，链表保持为空。firstmainNULL

为了解决这个问题，函数不应该将变量的副本传递给函数，而应该传递指向这些变量的指针（即引用）。因此，您应该将函数的参数更改为以下内容：mainfirstlastreadCarsreadCars

void readCars(char *fname, car *newCar, car **first, car **last)

您还应该将上面引用的行更改为以下内容：

if ( *first == NULL ) { 
    *first = newCar; // the new node is the only node so it is the first node
    *last = newCar; // the new node is the only node so it is also the last node
} else {
    (*last)->next = newCar; // the new node is next node of the previous last node
    (*last) = newCar; // the new node is now the last node
}

此外，函数的参数不作为参数使用。相反，您将其用作局部变量，因此您应该将其声明为局部变量并将其从函数的参数列表中删除。newCarreadCars

编辑：如下面的评论中指出的那样，您正在对函数和中的参数执行相同的操作。您也应该从这些函数中删除该参数。ptrprintCarsfreeMemory

我的解决方案是：

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

typedef struct car {
    char brand[15];
    int year; 
    struct car *prev; 
    struct car *next; 
} car;

void readCars(char *fname, car *newCar, car **first, car **last);
void printCars(car *ptr, car *first);
void freeMemory(car *ptr, car *first);

int main(int argc, char *argv[]) {
    car *first = NULL; // a pointer to the first node of the linked list
    car *last = NULL; // a pointer to the last node of the linked list
    car *newCar; // a pointer for new nodes
    car *ptr; // a pointer for iterating the linked list
    if(argc != 2) {
        printf("No filename provided.\n");
        exit(0);
    } 
    printf("Reading the file %s.\n", argv[1]);
    readCars(argv[1], newCar, &first, &last);
    printCars(ptr, first);
    freeMemory(ptr, first);
    printf("Program ended.\n");
    return(0);
}

void readCars(char *fname, car *newCar, car **first, car **last) {    
    FILE *tiedosto;
    char rivi[22];
    if ((tiedosto = fopen(fname, "r")) == NULL) {
        printf("Failed to open the file.\nProgram ended.\n");
        exit(0);
    }
    while (fgets(rivi, 22, tiedosto) != NULL) {
        if ((newCar = (car*)malloc(sizeof(car))) == NULL) { // allocate memory for a new node
                perror("Memory allocation failure.\n");
                exit(1);
        }
        sscanf(rivi, "%s %d", newCar->brand, &newCar->year); // set the  car brand and age to the new node
        newCar->next = NULL; // set the pointer to next node to NULL as  there is no next node
        newCar->prev = *last; // set the pointer to the previous node to  the previous last node (NULL if there was no previous)
        if (*first == NULL) {
            *first = newCar; // the new node is the only node so it is the  first node
            *last = newCar; // the new node is the only node so it is also  the last node
        } else {
            (*last)->next = newCar; // the new node is next node of the previous last node
            *last = newCar; // the new node is now the last node
        }
    }
    fclose(tiedosto);
    printf("File read into a linked list.\n");
}

void printCars(car *ptr, car *first) {
    if(first == NULL) {
        printf("No cars found.\n");
    } else {
        ptr = first;
        int count = 1;
        while (ptr != NULL) {
            printf("%d. car: %s from the year %d.\n", count, ptr->brand,  ptr->year);
            count += 1;
            ptr = ptr->next;
        }
    }
}

void freeMemory(car *ptr, car *first) { 
    ptr = first;
    while (ptr != NULL) {
        first = ptr->next;
        free(ptr);
        ptr = first;
    }
    printf("Memory freed.\n");
}

主要区别在于 readCars 需要一个指向第一个和最后一个的指针的指针。这样，当函数结束时，first 和 last 的值不会更改为初始值（在 NULL 中）。 如果您已经理解，请告诉我。