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Are you trying to debug? Got breakpoint in that line? What command are you using to run the exe from the IDE?
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thanks smith...
I got that problem out....Donno I was using a char pointer before...Now I changed it to array...strange but its working now....
I was debugging it by pressing f5 and by putting a break point
Thanks anyway
I am a learner Always
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I have a task where I am supposed to display the data(Clientname, id , year, license) on a list control from an XML file however when a new data is appended to the XML file with the same Clientname. I want the latest (i.e. recently appended data)should display.
Please let me know which function is required to perform this???
Thanks in advance.
modified on Wednesday, February 4, 2009 3:00 AM
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- Detecting a change to a file - you want to use directory change notifications[^].
- Displaying the last item in a list in an XML file? Kind of depends on the schema of the XML file, but you could use XPath to find the last item, or iterate through the file - there are a lot of techniques - I'd advise you to work out how to use the XML library you're using to do this.
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//Works
<br />
int py[5] = {0,1,2,3,4};<br />
int* pyp = (int*)py;
printf("%d",pyp[2]);<br />
<br />
//Doesn't.
<br />
show "2Dpointer\n";<br />
int py2[2][2] = {0,1,2,3};<br />
int **pyp2 = (int**)py2;<br />
<br />
printf("py2:%d",py2);<br />
printf("pyp2:%d",pyp2);<br />
<br />
printf("**pyp2[0][0]",pyp2[0][0]);
printf("**pyp2%d",**pyp2);
<code><br />
<br />
why? <br />
<br />
<div class="ForumSig">:beer:</div>
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All index [] notation is converted to pointer arithmetic.
So, you declare
int py2[2][2];
the compiler knows the total size (2*2 = 4) and how wide each row is (2).
So if you do
py2[1][0]
the compiler can convert to
py2 + (1*2) + 0
But then you do
int **pyp2 = py2;
Now you have created an alias and lost all information about the dimensions of the array.
So, when you try to do
pyp2[0][0]
the compiler can only get to
pyp2 + (0*?) + 0
it doesn't know what to fill in for ?
...cmk
The idea that I can be presented with a problem, set out to logically solve it with the tools at hand, and wind up with a program that could not be legally used because someone else followed the same logical steps some years ago and filed for a patent on it is horrifying.
- John Carmack
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but why does it work here ?
<br />
int py[5] = {0,1,2,3,4};<br />
int* pyp = (int*)py;
printf("%d",pyp[2]);<br />
As it shows, the memory has been laid out for these values,
[START_ADDRESS]4bytes,4bytes,4bytes,4bytes,4bytes.
So when you assign the pointer, it starts to point to the first item. Further when you make a pyp++ or pyp[], it's landing up on the right element. Here too, I'm not passing it any information about the size of the array. It's just the pointer and it's taking it up.
2D arrays, it's not too different from 1D. The next batch of numbers would be following.. err.. err. Okay as I'm typing this I'm getting it  ..I'm not giving it any clue where the first row ends.. well..is that what you mean? sounds convincing but why not it work atleast for the first item? I'm just saying **pyp2. Just trying to get the value of the element it's currently pointing to. Why even this doesn't work?
btw, why the complier is not warning me about a meaningless assignment?
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py2 is an address.
The contents of py2 are 0, 1, 2, 3.
Agreed ?
pyp2 is a pointer to a pointer.
pyp2 is initialized with the address of py2.
Now when you say *pyp2 it is taking the first value of 0 which it thinks is another address.
So, **pyp2 is trying to read the value at address 0. // Crash
If the contents of py2 were 5, 6, 7, 8,
**pyp2 will be trying to read the value at address 5. // Still Crash.
«_Superman_»
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So, **pyp2 is trying to read the value at address 0. // Crash
Why should it crash?
*pyp2 = first row.
**pyp2 = first element of the first row.
What's wrong with it?
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Smith# wrote: *pyp2 = first row.
**pyp2 = first element of the first row.
Wrong. I have tried to explain it in my previous post.
«_Superman_»
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Now when you say *pyp2 it is taking the first value of 0 which it thinks is another address.
Yes. that's the address of the first pointer. I called it as 1st row. With **pyp2, I try to access the first element. in other words, *pyp2[0].
I agree that you cannot ask it go to the next pointer since it doen't know where first one ends. But what's the problem to access the "first-most" element? We have the address, what's the problem with taking it's address? Am I still missing something?
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To access the first element you simply do *pyp2. Try it.
«_Superman_»
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That will give me the address of the first element. Not the value.
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It is giving me the value.
Try to understand that this is exactly why your code is crashing.
«_Superman_»
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Are you using VC6.0 Advanced Auto-Deferencing Compiler ? lol just kidding.
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py2 is an int *, not an int **. Extra dimensions of arrays DO NOT imply extra indirection (i.e. extra levels of pointer).
Therefore your C cast when assigning py2 to pyp2 is breaking the type safety that's trying to tell you what your mistake is.
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argh okay. accepted. is there way I can do it? I think it can be done by allocating a memory of that dimension dynamically and assigning the pointers one by one.bullshit. I meant assigning the pointers.
modified on Wednesday, February 4, 2009 4:02 AM
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Try initialising pyp2 like this:
int p1[2] = { 1, 2 };
int p2[2] = { 1, 2 };
int* pyp2[2] = { p1, p2 };
pyp2 is of type (int*)[2], which is equivalent to int **.
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hmm, so we cannot simply take a pointer and navigate through..  .. the complier wants it to be informed about these..
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I think I just ripped through mr.Smith
tried this :
<br />
int py2[2][2] = {10,11,12,13};<br />
<br />
int* sptr = &py2[0][0];<br />
int** dptr = &sptr;<br />
<br />
printf("\nAtlast:%d\n",**dptr);<br />
*dptr+=1;<br />
*dptr+=2;<br />
printf("\nAtlast:%d\n",**dptr);<br />
o/p:10,13
-Anderson.  .
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The key point is that arrays and pointers are NOT the same.
An array can be converted into a pointer and pointer arithmetic (and indexing) looks the same as array subscripting, but they aren't. They coincide only for linear indexing (that's the case of monodimensional arrays)
int **pyp2 is a "[pointer to a [pointer to [an int]]]"
The memory layout it expects is like
int** ----> [int*, int*, int*, ...]
| | |
| | |
| | V
| V [int,int,...]
V [int,int,int...]
[int, int, ...]
While int py2[2][2] is an "[array of [array of [int]]]"
The memory layout it expects is like
[[int,int],[int,int]]
That's completely different.
2 bugs found.
> recompile ...
65534 bugs found.
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As you know pointers & arrays both move consecutive. Then how come it be totally different? To some extent I can agree. Because the next pointer in the **ptr can be made to point to some other *p pointing somewhere. But here that's not the case.. it's array, quite consecutive and we are making our **ptr point to it. What do you think about the below approach ? getting the address of the first row pointer & getting the address of it and then deferencing.
<br />
int py2[2][2] = {10,11,12,13};<br />
<br />
int* sptr = &py2[0][0];<br />
int** dptr = &sptr; <br />
<br />
printf("\nAtlast:%d\n",**dptr); <br />
*dptr+=1;<br />
*dptr+=2;<br />
printf("\nAtlast:%d\n",**dptr);<br />
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It seems you're confusing two well distinct concepts (don't worry: it takes about three years to me to get away from that confusion...).
Let's reconsider your points:
Smith# wrote: As you know pointers & arrays both move consecutive
No. Pointers "move" (in th sense they can receive another address as their own stored value). Array just "are". Array indexes "move" (in the sense they can be used to indicate other array cells).
Smith# wrote: how come it be totally different
Because they ARE totally different. They simple have a "similar" external interface (same operators).If you don't get this point may be I was not able to explain the first point, but until you don't get it you cannot move away from your trivia. So loop over and over until you get it, find other sources than me (if you find my way to explain is not clear) but there is no way to move over until this crucial point is clear.
Smith# wrote: Because the next pointer in the **ptr can be made to point to some other *p pointing somewhere. But here that's not the case.. it's array,
You know, but compiler doesn't.
int** for the compiler is just 4 bytes of memory containing the address of an int* that is another 4 byte of memory, somewhere else, containing the address of an int.
Whatever you think about this is irrelevant to the compiler.
int[] is a consecutive bounce of integers at some place in memory. This place has an address.
When you do
int a[2] = { 0,1 };
int* pa = a;
you assign the address of the array head to the pointer.
When you do
*pa = 5;
the compiler assign 5 to the memory whose address is stored in "pa".
This accidentally is a[0] (but it is not something the compiler is aware of).
When you do
pa[1] = 6;
the compiler assign 6 to the memory whose address is given by pa+1*sizeof(int), that accidentally is a[1] (again you know, not the compiler: there is nothing in "pa" that makes the compiler aware of the fact that it is NOW pointing to an array and sometime later on to an individual integer and some later on to an integer inside a class or struct.
The design of the [] operator appllied to poitner or to array makes this operation giving a same result just because for monidimensional array, the same calculation is done.
Bidimensional array are different:
int a[2][2] = { 1,2,3,4 };
a[1,1] = 5;
means "set to 5 the cell "1" of the subarray "1" of the bidimensional array.
But
int **ppa = a;
ppa[1][1] = 8;
means: "take the value of "ppa", add 1*sizeof(int*) (note: int*, not int) fetch the value stored at such address, add 1*sizeof(int) (yes: now it's int) and this is the address where to store 8.
Now: since ppa contains the address of a[0][0] and sizeof(int*) is equal to sizeof(int) (at least on 32 bit Win-Intel), ppa[1][1] means:
take the address of a[0][0] add 1*sizeof(int*) (gives incidentally the address of a[1][0]), get the value (it's 2) add 1*sizeif(int) (gives 6) and store 8 to the memory whose address starts at "0x00000006" (6). That's out of your executable address space, hence the crash.
The compiler translate statements one by one without keeping memory of what a statement "means" being after another one.
int[][] is one thing, int** anoter, and the [] operator on a array is different to the [] operator on a pointer. Assigning an array address to a "pointer to pointer" doesn't change its nature of "pointer".
mondimensional array and simple pointer when applied to the [] operator give same results because in this degeneral case, the "resolution formula" is the same, since it start to be different from the second therm, that for one-only indirection, is not present.
2 bugs found.
> recompile ...
65534 bugs found.
modified on Monday, February 9, 2009 2:11 AM
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Hey Sorry man, I'm just checking your reply. Before even reading it I felt I should thank you for your helping effort. I'll read it, and if I find it reasonable, I'll give up the weapons & surrender  . Anyway thanks a lot.
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Now, let's move a little bit on:
If you agree with my previous post, consider this:
int a [2][2] = { 1,2,3,4 };
int[2]* ppa = a;
Do you get it? Instead of int**, I use int[2]*: a "pointer to an array of two integers". considering the int[][] as "array of 2 arrays of two integers".
The type of int** is int* (that doesn't exist in memory since we store arrays), but the value of int[2]* is int[2] that's just one row of the matrix ( in our case
{{1,2},{3,4}}
Now ppa[1][1] is "take the address stored in ppa, add 1*sizeof(int[2]). This is the int[2] representing the second row. Now we're anymore dealing with a pointer, but with an array, whose cell "1" has the value "4".
2 bugs found.
> recompile ...
65534 bugs found.
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