CODEWITHSUNDEEP
cpointers
C-Noob
C-Noob

Reputation: 31

Struggling with pointers in C

I thought I understood basic concepts of pointers as per C tutorials, but I really get confused when actually coding. I have some questions:

1 - Lets say I have:

customList arrayOfLists[3]; //global
//...
void someMethod()
{
   int i;
   for (i=0; i<3; i++)
   {
        customList l = arrayOfLists[i];

        //METHOD1
        //Next time I come back to this function, element is still there!
        removeFirstElement(&l);              

        //METHOD2
        //Next time I come back to thsi function, element is gone!
        //removeFirstElement(&(arrayOfLists[i])); 
   }
}

Why would both Method1 and Method2 not work the same way? In both cases I am basically saying remove the first element of the list located at addressX. Does it create a copy of the array?

2 - What is the difference between:

struct1.ptr1->ptr2->someIntValue //1
&(struct1).ptr1->ptr2->someIntValue //2

The 2nd way doesn't make sense to me and I don't really know what's going on there, but it seems to work. I was expecting the 1st way to work but it is not giving me the right answer.

3 - Lets say I do this:

ptr2 = ptrToStruct;
ptr1 = ptr2;
ptr2->intProp = 5;
ptr2 = ptrToStruct2;

Is ptr1 pointing to the initial memory location still, or is it the same ptr2? What is ptr1->intProp?

Thanks.

Upvotes: 3

Views: 232

Answers (5)

Tim Enghel
Tim Enghel

Reputation: 215

Pointers are hard to grasp because they are abstracting the memory of the computer without referring to that memory in the description of the operation you can do with pointers.

Pointers are much easier to understand if you learn a bit of assembler.

Upvotes: 1

Jeff Mercado
Jeff Mercado

Reputation: 134601

For question 1, method 1 is taking the address of the variable l. l is initialized to hold a copy the value of the i'th item of the array. This is a completely different memory location. Note that if the removeFirstElement() function depends on the the argument being a pointer to the array, it will fail miserably using method 1.

For question 2, line 2 is valid, but not for the reason you probably think. What it evaluates to is getting the address of struct1.ptr1->ptr2->someIntValue. Note that you did not take the address of the struct, but the member of the member of the member it was pointing to. You probably meant it to be as:

(&struct1).ptr1->ptr2->someIntValue

This would fail miserably.

The member access operator (.) is used on struct values to access a member. What you have on the LHS is a pointer to a struct and should be using the arrow operator (->) to access the member. Or, dereference the pointer first, then use the dot (what the arrow operator is equivalent to).

(*(&struct1)).ptr1
/* is equivalent to: */
(&struct1)->ptr1

For question 3, assuming ptr1 and ptr2 are pointers to whatever struct ptrToStruct is pointing to, then yes, ptr1 is still pointing to whatever it was pointing to. You never modified the ptr1 variable, only the struct that it was pointing to. Since both ptr1 and ptr2 are pointing to the same structure (in other words, aliased), you should see the same value when accessing the member ptr1->intProp.

Upvotes: 0

pmg
pmg

Reputation: 108986

  1. customList is a local pointer; arrayOfLists is a global array. They live in different memory regions.

  2. The & has lower precedence than . or ->. If I were to simplify the 1st expression to j, the 2nd expression would be simplified as &j.

  3. Pointers are just like any other type. In the code below

    int j, k;
k = 5;
j = k;
k = 12;

    does j have 5 or 12?

Upvotes: 0

Mahesh
Mahesh

Reputation: 34655

Q3)

ptr2 = ptrToStruct;
ptr1 = ptr2;
ptr2->intProp = 5;
ptr2 = ptrToStruct2;

Is ptr1 pointing to the initial memory location still, or is it the same ptr2?

Yes, ptr1 points to the memory location where ptrToStruct is pointing to. Affecting ptr2 to point to a different location (ptrToStruct2) doesn't affect ptr1.

What is ptr1->intProp?

5.

Upvotes: 0

Heisenbug
Heisenbug

Reputation: 39194

customList l = arrayOfLists[i];
//METHOD1
//Next time I come back to this function, element is still there!
removeFirstElement(&l);

This is wrong: you need something like:

customList *l = &arrayOfLists[i];
removeFirstElement(l);

For the second question: 

struct1.ptr1->ptr2->someIntValue // this will give you someIntValue
&(struct1).ptr1->ptr2->someIntValue // this will give you the address of memory where someIntValue is stored

Upvotes: 0

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