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[SOLVED] Defining Structure Array at Run-Time in C Language??

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Hello!!! Everyone i am using Ubuntu 14.04 and Code::Blocks GCC to do some programming in C language.

I want to write a program, in which i have to declare a structure array, the size is not fixed and i have to define it at run-time.

Here is my Code which is working Properly but i am not able to understand few points, hope someone will clarify my doubts.

Code C - [expand]
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#include <stdio.h>
#include <stdlib.h>
 
typedef struct
{
    unsigned char Name[10];
    unsigned int Fees;
}DATABASE;
 
int main()
{
    DATABASE *DataBase;
    unsigned int NumberOfStudents,i;
    printf("Enter the Number of Students?\n");
    scanf("%d",&NumberOfStudents);
 
    DataBase = (DATABASE *)malloc(sizeof(DATABASE)*NumberOfStudents);
    if(DataBase == NULL)
    {
        printf("\nUnsufficient Memory for Operation\n");
        exit(1);
    }
    printf("\n");
    for(i=0;i<NumberOfStudents;i++)
    {
        printf("\nEnter the Name of Student %d ?\n", i+1);
        scanf("%s",(DataBase+i)->Name);
        printf("\nEnter the Fees of Student %d ?\n", i+1);
        scanf("%u",&(DataBase+i)->Fees);
    }
 
    printf("\n");
    for(i=0;i<NumberOfStudents;i++)
    {
        printf("\nStudent %d Name is %s \n", i+1,(DataBase+i)->Name);
        printf("\nStudent %d Fees is %u \n", i+1,(DataBase+i)->Fees);
    }
    free(DataBase);
    return 0;
}


Code_Working.png


First of all tell me is it right method.
If yes then i want to know one more question

printf("\nEnter the Fees of Student %d ?\n", i+1);
scanf("%u",&(DataBase+i)->Fees);

I dont know why i write '&' in scanf statement, i know that in scanf one have to pass the address of the variable.

Does (DataBase+i)->Fees) doesn't points to array, and after adding & it will point to address.
Then why not with the Name String, because it is declared as an array and first byte of array points to address.

Please tell me if i am right or not.
 

in C function parameters are passed by value (a copy of the parameter is made and the function does have access to the original parameter). If a function needs to pass data to the original parameter one uses the & (address of operator) to pass the address of the parameter to the function which can then access and alter the memory allocated to the parameter. e.g.
Code: 
    printf("Enter the Number of Students?\n");
    scanf("%d",&NumberOfStudents);
the address of NumberOfStudents is passed to scanf() which can return the value read into the memory allocated to it

in the case of an array as a function parameter, e.g.
Code: 
        printf("\nEnter the Name of Student %d ?\n", i+1);
        scanf("%s",(DataBase+i)->Name);
there is no need to use & to take the address of (DataBase+i)->Name because it is already an address, i.e. the array name (DataBase+i)->Name is a constant pointer to the first element.
 
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