Question

我在初始化结构数组时遇到问题。我不确定我是否做得对，因为我得到“从不兼容的指针类型初始化”＆amp; “从不兼容的指针类型分配”。我在代码中添加了这些警告，当我尝试从结构中打印数据时，我只是得到垃圾，例如@@ ###

typedef struct
{
    char* firstName;
    char* lastName;
    int day;
    int month;
    int year;

}student;

//初始化数组

    student** students = malloc(sizeof(student));
    int x;
    for(x = 0; x < numStudents; x++)
    {
        //here I get: "assignment from incompatible pointer type" 
        students[x] = (struct student*)malloc(sizeof(student));
    }

    int arrayIndex = 0;

//添加struct

 //create student struct
        //here I get: "initialization from incompatible pointer type"
        student* newStudent = {"john", "smith", 1, 12, 1983};

        //add it to the array
        students[arrayIndex] = newStudent;
        arrayIndex++;

Answer 1

这是不正确的：

student** students = malloc(sizeof(student));

您不需要**。你需要一个*和足够的空间来容纳你需要的学生

student* students = malloc(numStudents * sizeof *students);
for (x = 0; x < numStudents; x++)
{
    students[x].firstName = "John"; /* or malloc and strcpy */
    students[x].lastName = "Smith"; /* or malloc and strcpy */
    students[x].day = 1;
    students[x].month = 12;
    students[x].year = 1983;
}

Answer 2

与编译器警告无关，但您的初始malloc是错误的;你想要的：

malloc(sizeof(student *)* numStudents)

为学生分配总共'numStudents'指针的空间。这一行：

students[x] = (struct student*)malloc(sizeof(student));

应该是：

students[x] = (student*)malloc(sizeof(student));

没有“结构学生”这样的东西。你已经声明了一个未命名的结构，并将它定义为'student'。比较和对比：

struct student
{
    char* firstName;
    char* lastName;
    int day;
    int month;
    int year;

};

这会创建一个'struct student'类型，但要求你（在C中）明确地引用struct student而不仅仅是其他地方的学生。对于C ++，此规则已更改，因此您的编译器可能会有点模糊。

至于：

student* newStudent = {"john", "smith", 1, 12, 1983};

应该是：

student newStudent = {"john", "smith", 1, 12, 1983};

由于大括号语法是直接文字，而不是你需要指向的其他地方。

编辑：经过反思，我认为aaa可能比我更了解这一点。是否有可能无意中在任何地方使用额外的指针解除引用？所以你想要：

student* students = malloc(sizeof(student) * numStudents);

/* no need for this stuff: */
/*int x;
for(x = 0; x < numStudents; x++)
{
    //here I get: "assignment from incompatible pointer type" 
    students[x] = (struct student*)malloc(sizeof(student));
}*/

int arrayIndex = 0;

和

student newStudent = {"john", "smith", 1, 12, 1983};

//add it to the array
students[arrayIndex] = newStudent;
arrayIndex++;

使数组不在newStudent范围之外使用。否则将指针复制到字符串是不正确的。

Answer 3

student* students = malloc(sizeof(student)*numStudents);
int x;
for(x = 0; x < numStudents; x++)
{
    student newStudent = {"john", "smith", 1, 12, 1983}; // string copy are wrong still
    students[x] = newStudent;
}

Answer 4

初始化时，不应该是这样吗？

student** students = (struct student**)malloc(sizeof(student*)*numStudents);

但是，为什么指针指针？只是用指向struct的指针就足够了。

C - 初始化结构数组

4 个答案: