c - 如何使用指针和动态内存将结构传递到函数中进行打印 [C 编程]

Question

我试图将一个结构传递到一个将打印内容的函数中。我正在寻求任何提示或建议。

结构体

struct student
{
    char *fname;
    char *lname;
    float score;
};

创建原型(prototype)

void printStudents(struct student **s, int student_size);

主要功能

int main()
{
    int base_number, i;
    int students;

    struct student *ptr;

    printf("Enter Number of Students: ");
    scanf("%d", &base_number);
    ptr = (struct student*)malloc(base_number * sizeof(struct student));
    students = base_number;

    printf("\nEnter Student Information\nExample: John Smith 98.50\n\n");

    for(i = 0; i < base_number; i++)
    {
        printf("Student %d: ", i+1);
        scanf("%s %s %f", &(ptr+i)->fname, &(ptr+i)->lname, &(ptr+i)->score);
    }

    printStudents(&ptr, students);

    return 0;
}

打印学生函数

void printStudents(struct student **s, int student_size)
{
    int i;
    struct student *temp = (struct student*)malloc(student_size * sizeof(struct student));

    for(i = 0; i < student_size; i++)
        printf("%s %s %.2f\n", (temp+i)->fname, (temp+i)->lname, (temp+i)->score);
}

Answer 1

1. 将指针传递给指针是否有任何要求？直接传ptr不行吗？

2. 为什么要构建另一个数组 temp？这只是浪费内存..

.

void printStudents(struct student *s, int student_size);

..

printStudents(ptr, students);

...

void printStudents(struct student *s, int student_size)
{
    int i;

    for(i = 0; i < student_size; i++)
        printf("%s %s %.2f\n", s[i].fname, s[i].lname, s[i].score);
}