Pointer Mock Exam Question

Hi,

Ive been given this code:

Code:

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int i=0, j=0, *k=0, *l=0;
k = l;
l = &i;
j = *k;

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Ive also been told that this code crashes. The question is, explain the cause of the crash, and correct the program by rearranging the istructions without modifiying them.

Can someone help me to understand why it is crashing. Ive been told that line 4 is the problem, but I dont really understand why.

Why not compile it and see what happens?

Ok. here's a hint. Try exchanging instructions 2 and 3 and see if it crashes now.

It doesnt crash, but I still cant see the problem. Im assuming it is due to the order in which things are being defined? Can k not be pointed to address 0?

I would think that pointer can only hold addresses or NUll which is 0 so...I think that this

statement

j = *k;

would be wrong...

Sorry , I was wrong...above

You get the error here

Code:

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k = l;

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because the pointers have not been initialized yet...

then as mentioned switching the two lines...

Code:

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l = &i;
k = l;

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Now initialization has happened thus is valid...

No, the error is in dereferencing a null pointer. There's nothing wrong with assigning NULL to a pointer.


Quzah.

Ive been looking at this again and I still dont understand why the error is occuring and its starting to bug me a bit.

Is it that since k is set equal to l on line 2, and at that point l is equal to 0 (setting k to 0 too), even when l is set equal to the address of i on the next line down, k is still set to 0 from line 2 and isnt affected by line 3.

Therefore when on line 4 when j is set equal to the pointer k, k is still being set equal to 0 memory address from line 2 and not infact being pointed to i?

Quote:

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Therefore when on line 4 when j is set equal to the pointer k, k is still being set equal to 0 memory address from line 2 and not infact being pointed to i?

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Yes. You observed correctly.
As quzah noted, the crash is caused because of line 4 where you dereference a null pointer.
Dereferencing a null pointer, results in undefined behavior which causes a segmentation fault and hence the crash.
Pointer (computing) - Wikipedia, the free encyclopedia

When the value of k is assigned to j, it contains zero. When you are dereferencing a pointer, you are getting the value at the address it contains. 0 is not valid address(Not in protected mode, at least) and that's your error.

In the same way, is this why reversing line two and three fixes the problem, because then l is pointing to location i before equating k to l, therefore giving this:

j = *k = l = i = 0

(0 = integer 0 not address 0)

Quote:

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Originally Posted by spadez

In the same way, is this why reversing line two and three fixes the problem, because then l is pointing to location i before equating k to l, therefore giving this:

j = *k = l = i = 0

(0 = integer 0 not address 0)

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Yes. I suggest using a debugger and trace out and see this for yourself.

Thank you very much for the help. Ive tried it in the debugger and it seems to confirm these results.

Would anyone mind explaining what the term "dereferencing a null pointer" actually means?

Quote:

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Originally Posted by spadez

Would anyone mind explaining what the term "dereferencing a null pointer" actually means?

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Suppose you have a pointer named p. *p is one way to dereference the pointer. So, if p is a null pointer, we can say that *p dereferences a null pointer. Another way would be to write say, p->foo, where foo is some member of the type that p is a pointer to.

I find this slightly confusing, isnt *p defining p as a pointer. In which case how is *p "dereferencing the pointer". I might need this dumbing down a little :s

Quote:

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Originally Posted by spadez

isnt *p defining p as a pointer.

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Not quite. Consider the syntax for defining a pointer and initialising it to be a null pointer:

Code:

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int *p = NULL;

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Notice that the type name int is part of the syntax. Compare it to this dereferencing of p to print what the pointer points to:

Code:

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printf("%d\n", *p);

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Now, the type name of the pointed to type is not part of the syntax.

Quote:

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Originally Posted by spadez

I find this slightly confusing, isnt *p defining p as a pointer. In which case how is *p "dereferencing the pointer". I might need this dumbing down a little :s

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Another reason I dislike T * syntax...
You can also define pointers as:
T* p;
Which might make it clearer.
Even clearer is:
T* p1;
T* p2;
Instead of:
T *p1, *p2;

p2 belongs to the same line as the declaration, hence it is also a declaration. Merely putting "*p" somewhere does not mean you are declaring a new variable since there's no type before it. Instead it's dereferencing the pointer.