// 调用函数这个写法，在输入第二个数据的时候，程序运行错了。在main中操作的写法是可以。
# include <stdio.h>
# include <malloc.h>

struct Student
{
    int age;
    float score;
    char name[10];
};

void InputStudent(struct Student ** st, int len)
{
    int i;
    *st = (struct Student *)malloc( len * sizeof(struct Student ) );

    for (i=0; i<len; ++i)
    {
        printf("请输入第%d个学生的信息\n", i+1);
        printf("age = ");
        scanf("%d", &((*st[i]).age));
        printf("score = ");
        scanf(" %f", &((*st[i]).score));
        printf("name = ");
        scanf("%s", &((*st[i]).name));

        getchar();
    }
}

int main(void)
{
    int len;
    int i;
    struct Student * pArr;

    printf("请输入要输入学生信息的个数：\n");
    scanf("%d", &len);
//    pArr = (struct Student *)malloc( len * sizeof(struct Student) );

    //输入学生信息
    InputStudent(&pArr, len);

/*    for (i=0; i<len; i++)
    {
        printf("请输入第%d个学生的信息\n", i+1);
        printf("age =");
        scanf("%d", &pArr[i].age);
        printf("score = ");
        scanf("%f", &pArr[i].score);
        printf("name = ");
        scanf("%s", &pArr[i].name);
    }
*/
    //输出学生信息

    for (i=0; i<len; i++)
    {
        printf("第%d个学生的信息是：\n", i+1);
        printf("age =%d\n",pArr[i].age);        
        printf("score = %f\n",pArr[i].score);        
        printf("name = %s\n", pArr[i].name);
    }

    return 0;
}

[ 本帖最后由 hyddpjh 于 2014-2-27 11:16 编辑 ]

用指针使结构体在函数中传递要怎么弄呢？我这种写法错在哪了,哪位大哥给我指点下？

# include <stdio.h>
# include <malloc.h>

struct Student
{
    int age;
    float score;
    char name[10];
};

void InputStudent(struct Student * st[], int len)
{
    int i;
    for (i=0; i<len; ++i)
    {
        printf("请输入第%d个学生的信息\n", i+1);
        printf("age = ");
        scanf("%d", &st[i]->age);
        printf("score = ");
        scanf(" %f", &st[i]->score);
        printf("name = ");
        scanf("%s", st[i]->name);
        getchar();
    }
}

int main(void)
{
    int len;
    int i;
    struct Student * pArr[256];
    printf("请输入要输入学生信息的个数(最多输入256个)：\n");
    scanf("%d", &len);
    //输入学生信息
    for(i=0;i<256;i++) pArr[i] = (struct Student *)malloc( sizeof(struct Student ) );
    InputStudent(pArr, len);
    //输出学生信息
    for (i=0; i<len; i++)
    {
        printf("第%d个学生的信息是：\n", i+1);
        printf("age =%d\n",pArr[i]->age);       
        printf("score = %f\n",pArr[i]->score);       
        printf("name = %s\n", pArr[i]->name);
    }
    for(i=0;i<256;i++) free( pArr[i]);
    return 0;
}

试试这个看是不是你想要的

指针传递过程中出错了    InputStudent(&pArr, len);  pArr存的是学生内容的地址  &pArr   是储存 pArr自己本身的地址   假设[2][0][0][0] 是学生内容的地址  那么[2][2][2][2]是pArr自己本身的地址  在 这个地址里面 放的内容的 [2][0][0][0][0] 学生的信息

//一直很苦恼，搞不定，就继续往下学了，学了链表以后，忽然开朗，修改了程序，终于搞明白了，运行正常。想明白了又感觉挺简单，呵呵。谢谢各位的解答提醒，谢谢了。
//修改程序如下：
# include <stdio.h>
# include <malloc.h>

struct Student
{
    int age;
    float score;
    char name[10];
};

struct Student *  InputStudent(void)
{
    int i, len;
    struct Student * pArr;
   
    printf("请输入要输入学生信息的个数：\n");
    scanf("%d", &len);
    pArr = (struct Student *)malloc( len * sizeof(struct Student ) );
   
    for (i=0; i<len; ++i)
    {
        printf("请输入第%d个学生的信息\n", i+1);
        printf("age = ");
        scanf("%d", &pArr[i].age);
        printf("score = ");
        scanf(" %f", &pArr[i].score);
        printf("name = ");
        scanf("%s", &pArr[i].name);
        
        //    pArr ++;
        //    getchar();
    }
   
    return (pArr);
}

int OutputStudent(struct Student * pArr )
{
    int len, i;
    struct Student * p;
   
    p = pArr;
   
    printf("请输入要输出学生信息的个数\n");
    scanf ("%d", &len);
   
    for(i=0; i<len; i++)
    {
        printf("第%d个学生信息为：\n", i+1);
        printf("age =%d\n", p[i].age);
        
        printf("score =%f\n", p[i].score);
        
        printf("name = %s\n", p[i].name);
        
    }
   
    return 0;
}

int main(void)
{
    struct Student * pArr;
   
    pArr = InputStudent( );
    OutputStudent(pArr );
   
    return 0;
}