为“结构**记录”的属性赋值

Question

我在理解 C 双指针概念时遇到了麻烦。本质上，我正在尝试编写代码来检查 record_1 是否尚未设置。如果没有，请设置它。如果已设置，我们将在第一条记录.next 指针中添加一个新的struct record newRecord。

我使用的是双指针，因为它是教授要求的

我尝试使用 firstRecord = malloc(sizeof(struct record*)); 没有任何运气，并尝试取消引用 firstRecord。

遍历位于addRecord 函数中的记录的while 循环也无法按预期工作，因为我不知道如何处理双指针。

struct record
{
    int                accountno;
    char               name[25];
    char               address[80];
    struct record*     next;
};

int addRecord (struct record ** firstRecord, int accountno, char name[], char address[])
{
    if (firstRecord == NULL)
    {
        // Segmentation Fault here
        // (*firstRecord)->accountno = accountno;
        // Assign the name to the newRecord
        // strcpy((*firstRecord)->name, name);
        // Assign the name to the newRecord
        // strcpy((*firstRecord)->address, address);
        // Initialize the next record to NULL
        // (*firstRecord)->next = NULL;
    }
    else
    {
        // Define a new struct record pointer named newRecord
        struct record newRecord;
        // Assign the accountno of newRecord
        newRecord.accountno = accountno;
        // Assign the name to the newRecord
        strcpy(newRecord.name, name);
        // Assign the address to the newRecord
        strcpy(newRecord.address, address);
        // Initialize the next record to NULL
        newRecord.next = NULL;
        // Create a new record and add it to the end of the database
        struct record ** iterator = firstRecord;
        // Iterate through the records until we reach the end
        while (iterator != NULL)
        {
            // Advance to the next record
            *iterator = (*iterator)->next;
        }
        // Assign the address of newRecord to the iterator.next property
        (*iterator)->next = &newRecord;
    }

    return 1;
}

int main() {
    struct record ** firstRecord;
    firstRecord = NULL;

    addRecord(firstRecord, 1, "Foo", "Bar");
    addRecord(firstRecord, 2, "Foo", "Bar");

    return 0;
}

Answer 1

这不仅是教授要求的，也是你的申请要求的。您想要分配内存，并设置一个在您的函数外部定义的指针指向该内存。所以很自然，您需要引用该指针。而 C 允许您通过指向指针的指针来做到这一点。

所以你希望你的调用代码看起来像这样：

struct record * firstRecord = NULL;
addRecord(&firstRecord, 1, "Foo", "Bar");
addRecord(&firstRecord, 2, "Foo", "Bar");

您传递常规指针的地址，以便addRecord 可以写入它。它通过取消引用它的论点来做到这一点，如下所示：

int addRecord (struct record ** pFirstRecord, /* ... */ )
{
    if (pFirstRecord == NULL)
      return 0; // We weren't passed a valid address of a pointer to modify

    if(*pFirstRecord == NULL)
    {
      // Here we check if the pointed to pointer already points to anything.
      // If it doesn't, then proceed with adding the first record
    }
}

Answer 2

你不需要双指针，一个简单的指针就足够了。首先，您需要为新记录动态分配内存，然后为其设置值：

int main() {
    struct record *record1;
    record1 = NULL;

    if (record1 == NULL)
    {
        printf("\nrecord_1 is NULL\n");


        //dynamically allocate memory for record1
        record1 = malloc(sizeof(struct record);
        if (record_1 == NULL)
        {
            printf("Error allocating memory\n");
            return -1;
        }

        //set values to record1
        record1->accountno = ...;
        strcpy(record1->name, ...);
        strcpy(record1->address, ...);
        record1->next = NULL;
        return 0;
    }
    else
    {
        //do the same for record1->next
        record1->next = malloc(sizeof(struct record);
        if (record1->next == NULL)
        {
            printf("Error allocating memory\n");
            return -1;
        }

        //set values to record1
        record1->next->accountno = ...;
        strcpy(record1->next->name, ...);
        strcpy(record1->next->address, ...);
        record1->next->next = NULL;
        return 0;
    }
}

但请注意，在编写此程序的方式中，永远不会到达 else 部分，因为 record1 始终初始化为 NULL 并且没有任何形式的迭代。

您可以阅读this link 作为结构及其内存分配的参考。