c - 如何在 struct 中为 char* 字段分配内存

Question

如何为结构中的 char * 字段分配内存？

我的结构:

   struct student{
    int score;
    char* name;
    char* surname;
};
int main(){
    struct student st[];
    int i;
    int n = 5;
    for(i = 0; i < n; i++){
        printf("Score: \n");
        scanf("%d", &st[i].score);
        printf("Name \n");
        scanf("%s", &st[i].name);
        printf("Surname \n");
        scanf("%s",&st[i].surname)
    }

}

如何 malloc 到 char* name 和 char* surname ？
我必须有一个结构数组，格式为struct student st[]。

我不知道，如何合理地做到这一点。

void initialise_student( struct student *st, char* name, char* surname)
    {
     st->name = ( strlen( name ) + 1);
     st->surname = (strlen( surname ) +1 );
    }
    int main(){
     int i;
     int n = 5;
     struct student *st[n] = initialise_student();
     for(i = 0; i < n; i++){
       printf("Score: \n");
       scanf("%d", &st[i].score);
       printf("Name \n");
       scanf("%s", &st[i].name);
       printf("Surname \n");
       scanf("%s",&st[i].surname);
      }

如何匹配这个？

Answer 1

例如

#include <stdlib.h>
#include <string.h>

//...

struct student st[1];
char *name = "Marek";
char *surname = "Piszczaniuk";

st[0].name = malloc( strlen( name ) + 1 );
strcpy( st[0].name, name );

st[0].surname = malloc( strlen( surname ) + 1 );
strcpy( st[0].surname, surname );

st[0].score = 100;

您可以编写单独的函数来设置数组元素的数据成员名称和姓氏。

例如

_Bool set_name( struct student *st, const char *name )
{
    st->name = malloc( strlen( name ) + 1 );
    _Bool  success = st->name != NULL;

    if ( success )
    {
        strcpy( st->name, name );
    }

    return success;
}