Understanding dynamic memory allocation

Question

I am a beginner in C programming. I was recently taught how to use malloc, but I don't think that I quite understand it. Like why does it need a void * or any typecast as a matter of fact? Why does the syntax itself have (void *) in void *malloc(size_t size). And how does the variable assigned the malloc function know where the memory block begins from? Does the malloc function return an address or something after it has assigned a memory block?

In the class our prof gave us this program. I understand how 2d memory allocation works too.

#include<stdio.h>
#include<conio.h>
void main(void)
{
    int *studentInfo=NULL,i=1,j=10,k=0,l=0;
    //int *studInfo[10];
    int memLoc=0;
    clrscr();

    printf("How many Student Information You want to store:");
    scanf("%d",&j);
    printf("How many Subject Marks per student You want to store:");
    scanf("%d",&k);
    studentInfo=(int *)malloc(j*k*sizeof(int));

    //memLoc=0;
    for(l=0;l<j;l++)
    {
        printf("Enter Marks for %dth Student",l+1);
        for(i=0;i<k;i++)
        {
            printf("\nEnter Marks for Subject %d:",i+1);
            scanf("%d",studentInfo+(l*j)+i);
        }
    }
    //3 students and 3 subjects
    //memory allocated=3*3*2=18
    //0,1,2 student 0*no of students
    //3 4 5 student 1
    //6 7 8 student 2

    printf("\nInformation you Entered\n");
    for(l=0;l<j;l++)
    {
        printf("Makrs of Student %d:",l+1);
        for(i=0;i<k;i++)
            printf("\t%d",*(studentInfo+(l*j)+i));
        printf("\n");
    }

    //*(studentInfo)=10;
    //*(studentInfo+1)=20;
    //*(studentInfo+2)=30;
    //printf("%d\n",sizeof(studentInfo));
    //printf("%d\n",*(studentInfo));
    //printf("%d\n",*(studentInfo+i++));
    //printf("%d\n",*(studentInfo+i++));


    free(studentInfo);
    getch();
}

In this we are assigning the studentInfo pointer the malloc function right? So... how does studentInfo know that the address of the memory block is USA and not Antarctica?? And I know that it is not a good practise to typecast malloc with some other datatype. But why (int *). Why does malloc need a pointer?? If malloc needs a pointer that means that it is returning an address right? I asked this to my friend and he said no malloc doesn't return anything. And one more thing is it necessary that we need the typecast to be in brackets?

Please explain in very simple terms. Thank you.

Answer 1

Why does the syntax itself have (void *) in void *malloc(size_t size).

Because malloc is a function and functions have a return type. This tells that this function returns this particular type. So (void *) means malloc return a void *.

Does the malloc function return an address or something after it has assigned a memory block?

Malloc allocates the memory of the size specified into the heap and returns a pointer to that allocated memory.

In this we are assigning the studentInfo pointer the malloc function right? So... how does studentInfo know that the address of the memory block is USA and not Antarctica??

Not exactly. studentInfo is not assigned the malloc function but studentInfo points to the pointer returned by malloc. This was studentInfo now points to the allocated memory.

Why does malloc need a pointer??

malloc doesn't need a pointer. It takes the amount of memory you want to allocate as an argument.

You don't need to typecast here, as void * is automatically promoted.

And one more thing is it necessary that we need the typecast to be in brackets?

Yes, that's the syntax for typecasting. Without brackets, it will result in compilation errors.