What happens in memory during DMA using pointers?

[shivajikobardan]

Homework Statement

pointers in c

Relevant Equations

none

#include<stdio.h>
#include<stdlib.h>

int main()
{
    int *p;
    p=(int *) malloc(4);
    if(p==NULL)
    {
        printf("Insufficient memory");
        return;
    }

    printf("Enter a number\n");
    scanf("%d",p);
    printf("Value of p=%d",*p);
}

1) What happens in background when we do int *p? Can you show a figure about what it does in memory?

2) What happens in memory "p=(int *) malloc(4);" when we run this line? In surface, I've rote memorized, it allocates 4 bytes of memory, but I don't know in detail what happens in background.

3) printf("Enter a number\n");
scanf("%d",p);

I need to store a number in address of p, so should not I do &p instead of just p?

4) Value of p=p isn't it? Why *p? I've again rote memorized *p gives value of p, but I'm not clear about what's going inside memory during this all.

I'm asking these questions knowing I'm wrong because I've tested the code in codeblocks already. It's not to get the correct answer but a way to think to get a correct answer.

If we'd done;
int x=10;
int *p=&x;
I'd understand what's being done here.
It means content of p is initialzed with address of x. Like this:

but I don't understand the above case that I mentioned in question where we are only doing int *p. I find it meaningless. p is a pointer variable. But it contains whose address?

 

[Filip Larsen]

You may want to read up on stack vs heap allocation of variables in general.

p is a pointer variable. But it contains whose address?

p is assigned the result from the call to the malloc function, so if you know (look up) what malloc do you should be able to answer your own question.

 

[Mark44]

1) What happens in background when we do int *p? Can you show a figure about what it does in memory?

Uninitialized storage for p is allocated on the stack, storage large enough to hold a memory address. The amount of storage allocated depends on whether you're building your program in 32-bit mode or in 64-bit mode. Uninitialized means that the value stored there is whatever happens to already be in memory at that location. It's allocated on the stack because it's a variable that is declared inside a function (i.e., inside main()). If it had been declared outside of any function, it would have been allocated in static memory.

Your figure is reasonably accurate regarding the relationship between the p and x variables. The topic of pointers is one that students often have difficulties in understanding. I've taught classes in C many, many times, and I always present the topic of pointers using a diagram similar to yours. Having a diagram is very helpful.

2) What happens in memory "p=(int *) malloc(4);" when we run this line? In surface, I've rote memorized, it allocates 4 bytes of memory, but I don't know in detail what happens in background.

The malloc() function allocates 4 bytes of storage on the heap and returns the address of the first byte. This address is cast as a pointer to int and is stored in p. In your drawing, the address of this heap storage is 1000H (usually written as 0x1000 for hexadecimal numbers). This value is stored in the location identified as p. In your drawing, p's address is 0x2000.d

3) printf("Enter a number\n");
scanf("%d",p);

I need to store a number in address of p, so should not I do &p instead of just p?

The scanf() function's second argument here needs to be the address of the memory that will receive a value. p is a pointer to type int, so it already is an address.

4) Value of p=p isn't it? Why *p? I've again rote memorized *p gives value of p, but I'm not clear about what's going inside memory during this all.

Of course the value of a variable is its value, but that's always true and doesn't tell you anything. What your program intends to do is to print the value that is pointed to by p, not the value in p.
This line of code in your program, printf("Value of p=%d",*p);, will display the value you entered when scanf() was executed. That will be the value in the hypothetical address 0x1000.

 

[shivajikobardan]

Thank you all for your replies.