了解动态 memory 分配

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.

Answer 2

malloc returns void* because malloc allocates not only int , it can allocate floats, doubles even structures etc. Reason of returning pointer is because it can allocate so much memory in heap, heap is larger than stack and if allocated memory is so big for stack you will receive an error(probably stackoverflow) + copy memory is meaningless if we just can copy its address and refer it by pointer. And void * is a special kind that everything can be assigned no matter what type of address it is. Yes you need to cast it, if you don't then compiler will give you an error

Answer 3

The malloc function call allocates a block of memory large enough to hold j*k int objects and returns the address of the first element of that block - that address value is assigned to studentInfo , giving us something like this:

             int *       int
             +–––+       +–––+
studentInfo: |   | ––––> |   |
             +–––+       +–––+
                         |   |
                         +–––+
                          ...

Thus, studentInfo points to (stores the address of) the first int object in that sequence.

The (int *) preceding the malloc call in your code is a cast - it means "treat the result of the following expression (the value returned by malloc ) as an int * , or pointer to int ."

There only two circumstances where you must cast the result of malloc :

• you're compiling the code as C++;
• you're using an ancient K&R C compiler;

Unlike C, C++ does not allow direct assignment of void * to other pointer types.

Also, prior to the C89 standard, malloc returned char * , so a cast was required to assign the result to any other pointer type.

Otherwise, it's generally preferred to not cast the result - it can lead to maintenance headaches down the line. It would be a bit cleaner to write it as

studentInfo = malloc( sizeof *studentInfo * j * k );

Since the type of studentInfo is int * , the type of the expression *studentInfo is int , so the result of sizeof *studentInfo is the same as sizeof (int) .

Remember that sizeof is an operator , not a function - parens are only required if the operand is a type name.