Strange programming problems as of now..As you can see below i have assigned intFrontPtr to point to the first cell in the array. And intBackPtr to point to the last cell in the array...:
bool quack::popFront( int &popFront )
{
//items[count-1].n = { 9,4,3,2,1,0 };
nPopFront = items[0].n;
if ( count >= maxSize ) return false;
else
{
items[0].n = nPopFront;
intFrontPtr = &items[0].n;
intBackPtr = &items[count-1].n;
}
for (int temp; intFrontPtr < intBackPtr ;)
{
intFrontPtr++;
temp = *intFrontPtr;
*intFrontPtr = temp;
}
--count;
return true;
}
Its just my implementation of a cross between a queue and a stack..PopFront is a public method of the class object quack..The items is a private struct type 'item', it is within the quack.h. It has one member, 'int n'..But, that is irrelevent.
the comment in the code is the contents of my integer array, 'items'.
I am trying to Pop elements off the front of my array. WHat im thinking is that after i get the first item, i'll just incrememnt the frontPtr and transfer the item i got previously to the frontPtr i incremented!...
I cannot, for any reason, use a + or - shift by 1 or the use of stls, boosts, std's and the like..
Can someone help me with my homework assignment?
My suggestion are :
1). Put statement --count where it keeps object's state valid on exceptional condition.
2). clear your concepts of pointers which will help you a lot.
Generally, I would not recommend using an array if you want to pop items off the front. You would be much better off using a linked list, or some other structure which will allow you to remove items in O(1) time. It will be much easier to remove (pop) items this way.
As for your current code, I really can't comment without a better idea of what your class looks like. Please post the class definition at least so we can tell what all your variables are referring to, and what you're code is actually doing.
There are quite a few problems. But I believe your major one is this loop:
for (int temp; intFrontPtr < intBackPtr ;)
{
intFrontPtr++;
temp = *intFrontPtr;
*intFrontPtr = temp;
}
It looks to me like you are trying to shift all your items down. Since this is homework, I'm not going to give you the answer but I'll give you some hints to help debug this.
What you want to do is examine your items array at the beginning and end of the loop.
for (int temp; intFrontPtr < intBackPtr ;)
{
// whats does array look like here
intFrontPtr++;
temp = *intFrontPtr;
*intFrontPtr = temp;
// and what does array look like here
}
If you don't have experience with a debugger, add a function call that will dump your array.
Related
I am trying to insert data into a leaf node (an array) of a B-Tree. Here is the code I have so far:
void LeafNode::insertCorrectPosLeaf(int num)
{
for (int pos=count; pos>=0; pos--) // goes through values in leaf node
{
if (num < values[pos-1]) // if inserting num < previous value in leaf node
{continue;} // conitnue searching for correct place
else // if inserting num >= previous value in leaf node
{
values[pos] = num; // inserts in position
break;
}
}
count++;
} // insertCorrectPos()
Before the line values[pos] = num, I think need to write some code that shifts the existing data instead of overwriting it. I am trying to use memmove but have a question about it. Its third parameter is the number of bytes to copy. If I am moving a single int on a 64 bit machine, does this mean I would put a "4" here? If I am going about this completely wrong any any help would be greatly appreciated. Thanks
The easiest way (and probably the most efficient) would be to use one of the standard libraries predefined structures to implement "values". I suggest either list or vector. This is because both list and vector has an insert function that does it for you. I suggest the vector class specifically is because it has the same kind of interface that an array has. However, if you want to optimize for speed of this action specifically, then I would suggest the list class because of the way it is implemented.
If you would rather to it the hard way, then here goes...
First, you need to make sure that you have the space to work in. You can either allocate dynamically:
int *values = new int[size];
or statically
int values[MAX_SIZE];
If you allocate statically, then you need to make sure that MAX_SIZE is some gigantic value that you will never ever exceed. Furthermore, you need to check the actual size of the array against the amount of allocated space every time you add an element.
if (size < MAX_SIZE-1)
{
// add an element
size++;
}
If you allocate dynamically, then you need to reallocate the whole array every time you add an element.
int *temp = new int[size+1];
for (int i = 0; i < size; i++)
temp[i] = values[i];
delete [] values;
values = temp;
temp = NULL;
// add the element
size++;
When you insert a new value, you need to shift every value over.
int temp = 0;
for (i = 0; i < size+1; i++)
{
if (values[i] > num || i == size)
{
temp = values[i];
values[i] = num;
num = temp;
}
}
Keep in mind that this is not at all optimized. A truly magical implementation would combine the two allocation strategies by dynamically allocating more space than you need, then growing the array by blocks when you run out of space. This is exactly what the vector implementation does.
The list implementation uses a linked list which has O(1) time for inserting a value because of it's structure. However, it is much less space inefficient and has O(n) time for accessing an element at location n.
Also, this code was written on the fly... be careful when using it. There might be a weird edge case that I am missing in the last code segment.
Cheers!
Ned
I have a vector holding 10 items (all of the same class for simplicity call it 'a'). What I want to do is to check that 'A' isn't either a) hiding the walls or b) hiding another 'A'. I have a collisions function that does this.
The idea is simply to have this looping class go though and move 'A' to the next position, if that potion is causing a collision then it needs to give itself a new random position on the screen. Because the screen is small, there is a good chance that the element will be put onto of another one (or on top of the wall etc). The logic of the code works well in my head - but debugging the code the object just gets stuck in the loop, and stay in the same position. 'A' is supposed to move about the screen, but it stays still!
When I comment out the Do while loop, and move the 'MoveObject()' Function up the code works perfectly the 'A's are moving about the screen. It is just when I try and add the extra functionality to it is when it doesn't work.
void Board::Loop(void){
//Display the postion of that Element.
for (unsigned int i = 0; i <= 10; ++i){
do {
if (checkCollisions(i)==true){
moveObject(i);
}
else{
objects[i]->ResetPostion();
}
}
while (checkCollisions(i) == false);
objects[i]->SetPosition(objects[i]->getXDir(),objects[i]->getYDir());
}
}
The class below is the collision detection. This I will expand later.
bool Board::checkCollisions(int index){
char boundry = map[objects[index]->getXDir()][objects[index]->getYDir()];
//There has been no collisions - therefore don't change anything
if(boundry == SYMBOL_EMPTY){
return false;
}
else{
return true;
}
}
Any help would be much appreciated. I will buy you a virtual beer :-)
Thanks
Edit:
ResetPostion -> this will give the element A a random position on the screen
moveObject -> this will look at the direction of the object and adjust the x and Y cord's appropriately.
I guess you need: do { ...
... } while (checkCollisions(i));
Also, if you have 10 elements, then i = 0; i < 10; i++
And btw. don't write if (something == true), simply if (something) or if (!something)
for (unsigned int i = 0; i <= 10; ++i){
is wrong because that's a loop for eleven items, use
for (unsigned int i = 0; i < 10; ++i){
instead.
You don't define what 'doesn't work' means, so that's all the help I can give for now.
There seems to be a lot of confusion here over basic language structure and logic flow. Writing a few very simple test apps that exercise different language features will probably help you a lot. (So will a step-thru debugger, if you have one)
do/while() is a fairly advanced feature that some people spend whole careers never using, see: do...while vs while
I recommend getting a solid foundation with while and if/else before even using for. Your first look at do should be when you've just finished a while or for loop and realize you could save a mountain of duplicate initialization code if you just changed the order of execution a bit. (Personally I don't even use do for that any more, I just use an iterator with while(true)/break since it lets me pre and post code all within a single loop)
I think this simplifies what you're trying to accomplish:
void Board::Loop(void) {
//Display the postion of that Element.
for (unsigned int i = 0; i < 10; ++i) {
while(IsGoingToCollide(i)) //check is first, do while doesn't make sense
objects[i]->ResetPosition();
moveObject(i); //same as ->SetPosition(XDir, YDir)?
//either explain difference or remove one or the other
}
}
This function name seems ambiguous to me:
bool Board::checkCollisions(int index) {
I'd recommend changing it to:
// returns true if moving to next position (based on inertia) will
// cause overlap with any other object's or structure's current location
bool Board::IsGoingToCollide(int index) {
In contrast checkCollisions() could also mean:
// returns true if there is no overlap between this object's
// current location and any other object's or structure's current location
bool Board::DidntCollide(int index) {
Final note: Double check that ->ResetPosition() puts things inside the boundaries.
Okay, I have been set with the task of comparing this list of Photons using one method (IU) and comparing it with another (TSP). I need to take the first IU photon and compare distances with all of the TSP photons, find the smallest distance, and "pair" them (i.e. set them both in arrays with the same index). Then, I need to take the next photon in the IU list, and compare it to all of the TSP photons, minus the one that was chosen already.
I know I need to use a Boolean array of sorts, with keeping a counter. I can't seem to logic it out entirely.
The code below is NOT standard C++ syntax, as it is written to interact with ROOT (CERN data analysis software).
If you have any questions with the syntax to better understand the code, please ask. I'll happily answer.
I have the arrays and variables declared already. The types that you see are called EEmcParticleCandidate and that's a type that reads from a tree of information, and I have a whole set of classes and headers that tell that how to behave.
Thanks.
Bool_t used[2];
if (num[0]==2 && num[1]==2) {
TIter photonIterIU(mPhotonArray[0]);
while(IU_photon=(EEmcParticleCandidate_t*)photonIterIU.Next()){
if (IU_photon->E > thresh2) {
distMin=1000.0;
index = 0;
IU_PhotonArray[index] = IU_photon;
TIter photonIterTSP(mPhotonArray[1]);
while(TSP_photon=(EEmcParticleCandidate_t*)photonIterTSP.Next()) {
if (TSP_photon->E > thresh2) {
Float_t Xpos_IU = IU_photon->position.fX;
Float_t Ypos_IU = IU_photon->position.fY;
Float_t Xpos_TSP = TSP_photon->position.fX;
Float_t Ypos_TSP = TSP_photon->position.fY;
distance_1 = find distance //formula didnt fit here //
if (distance_1 < distMin){
distMin = distance_1;;
for (Int_t i=0;i<2;i++){
used[i] = false;
} //for
used[index] = true;
TSP_PhotonArray[index] = TSP_photon;
index++;
} //if
} //if thresh
} // while TSP
} //if thresh
} // while IU
Thats all I have at the moment... work in progress, I realize all of the braces aren't closed. This is just a simple logic question.
This may take a few iterations.
As a particle physicist, you should understand the importance of breaking things down into their component parts. Let's start with iterating over all TSP photons. It looks as if the relevant code is here:
TIter photonIterTSP(mPhotonArray[1]);
while(TSP_photon=(EEmcParticleCandidate_t*)photonIterTSP.Next()) {
...
if(a certain condition is met)
TSP_PhotonArray[index] = TSP_photon;
}
So TSP_photon is a pointer, you will be copying it into the array TSP_PhotonArray (if the energy of the photon exceeds a fixed threshold), and you go to a lot of trouble keeping track of which pointers have already been so copied. There is a better way, but for now let's just consider the problem of finding the best match:
distMin=1000.0;
while(TSP_photon= ... ) {
distance_1 = compute_distance_somehow();
if (distance_1 < distMin) {
distMin = distance_1;
TSP_PhotonArray[index] = TSP_photon; // <-- BAD
index++; // <-- VERY BAD
}
}
This is wrong. Suppose you find a TSP_photon with the smallest distance yet seen. You haven't yet checked all TSP photons, so this might not be the best, but you store the pointer anyway, and increment the index. Then if you find another match that's even better, you'll store that one too. Conceptually, it should be something like this:
distMin=1000.0;
best_photon_yet = NULL;
while(TSP_photon= ... ) {
distance_1 = compute_distance_somehow();
if (distance_1 < distMin) {
distMin = distance_1;
best_pointer_yet = TSP_photon;
}
}
// We've now finished searching the whole list of TSP photons.
TSP_PhotonArray[index] = best_photon_yet;
index++;
Post a comment to this answer, telling me if this makes sense; if so, we can proceed, if not, I'll try to clarify.
Trying not to lose it here. As you can see below I have assigned intFrontPtr to point to the first cell in the array. And intBackPtr to point to the last cell in the array...:
bool quack::popFront(int& nPopFront)
{
nPopFront = items[top+1].n;
if ( count >= maxSize ) return false;
else
{
items[0].n = nPopFront;
intFrontPtr = &items[0].n;
intBackPtr = &items[count-1].n;
}
for (int temp; intFrontPtr < intBackPtr ;)
{
++intFrontPtr;
temp = *intFrontPtr;
*intFrontPtr = temp;
}
return true;
}
In the else statement I'm simply reassigning to ensure that my ptrs are where I want them. For some reason I'm popping off the back instead of off the front.
Anyone care to explain?
I'm not entirely sure I understand what you're trying to do, but if I;m guessing right you're trying to 'pop' the 1st element of the array (items[0]) into the nPopFront int reference, then move all the subsequent elements of the array over by one so that the 1st element is replaced by the 2nd, the 2nd by the 3rd, and so on. After this operation, the array will contain one less total number of elements.
Not having the full declaration of the quack class makes most of the following guesswork, but here goes:
I'm assuming that item[0] represents the 'front' of your array (so it's the element you want 'popped').
I'm also assuming that 'count` is the number of valid elements (so item[count-1] is the last valid element, or the 'back' of the array).
Given these assumptions, I'm honestly not sure what top is supposed to represent (so I might be entirely wrong on these guesses).
Problem #1: your nPopFront assignment is reversed, it should be:
nPopFront = items[0].n;
Problem #2; your for loop is a big no-op. It walks through the array assigning elements back to their original location. I think you want it to look more like:
for (int i = 1; i < count; ++i)
{
items[i-1].n = items[i].n; // move elements from back to front
}
Finally, you'll want to adjust count (and probably top - if you need it at all) before you return to adjust the new number of elements in the data structure. The whole thing might look like:
bool quack::popFront(int& nPopFront)
{
if ( count >= maxSize ) return false;
if ( count == 0 ) return false; // nothing to pop
nPopFront = items[0].n;
intFrontPtr = &items[0].n; // do we really need to maintain these pointers?
intBackPtr = &items[count-1].n;
for (int i = 1; i < count; ++i)
{
items[i-1].n = items[i].n; // move elements from back to front
}
count -= 1; // one less item in the array
return true;
}
The original question seems to be that you don't understand why the function popFront returns 3 times when there are 3 elements?
If that's the case, I think you are missing the point of recursion.
When you make a recursive call, you are calling the same function again, basically creating a new stack frame and jumping back to the same function. So if there are 3 elements, it will recurse by encountering the first element, encountering the second element, encountering the third element, returning from the third encounter, returning from the second encounter, and returning from the first encounter (assuming you are properly consuming your array, which you don't appear to be).
The current function cannot return until the recursive call has iterated, thus it may appear to return from the last element before the second, and the second before the first.
That is how recursion works.
I wasn't able to make sense of your example, so I whipped one up real fast:
#include <iostream>
using namespace std;
bool popfront(int* ptr_, int* back_) {
cerr << ptr_[0] << endl;
if(ptr_ != back_) {
popfront(++ptr_, back_);
}
return true;
}
int main() {
int ar[4] = {4,3,2,1};
popfront(ar, ar + 3);
return 0;
}
That's not great, but it should get the point across.
Can't you just use a std::list?
That makes it really to pop from either end using pop_front or pop_back. You can also add to the front and the back. It also has the advantage that after popping from the front (or even removing from the middle of the list) you don't have to shift anything around (The link is simply removed) which makes it much more efficient than what you are, seemingly, proposing.
I'm assuming you're trying to assign the popped value to nPopFront?
bool stack::popFront(int& nPopFront)
{
//items[4] = {4,3,2,1}
if ( intFrontPtr < intBackPtr )
{
nPopFront = *intFrontPtr;
++intFrontPtr;
}
return true;
}
bool quack::popFront(int& nPopFront)
{
if(items.n==0) throw WhateverYouUseToSignalError;
nPopFront = items[0];
for (int =0;i<items.n-1,++i){
items[i]=items[i+1]
}
//update size of items array
}
I thought i'd post a little of my homework assignment. Im so lost in it. I just have to be really efficient. Without using any stls, boosts and the like. By this post, I was hoping that someone could help me figure it out.
bool stack::pushFront(const int nPushFront)
{
if ( count == maxSize ) // indicates a full array
{
return false;
}
else if ( count <= 0 )
{
count++;
items[top+1].n = nPushFront;
return true;
}
++count;
for ( int i = 0; i < count - 1; i++ )
{
intBackPtr = intFrontPtr;
intBackPtr++;
*intBackPtr = *intFrontPtr;
}
items[top+1].n = nPushFront;
return true;
}
I just cannot figure out for the life of me to do this correctly! I hope im doing this right, what with the pointers and all
int *intFrontPtr = &items[0].n;
int *intBackPtr = &items[capacity-1].n;
Im trying to think of this pushFront method like shifting an array to the right by 'n' units...I can only seem to do that in an array that is full. Can someone out their please help me?
Firstly, I'm not sure why you have the line else if ( count <= 0 ) - the count of items in your stack should never be below 0.
Usually, you would implement a stack not by pushing to the front, but pushing and popping from the back. So rather than moving everything along, as it looks like you're doing, just store a pointer to where the last element is, and insert just after that, and pop from there. When you push, just increment that pointer, and when you pop, decrement it (you don't even have to delete it). If that pointer is at the end of your array, you're full (so you don't even have to store a count value). And if it's at the start, then it's empty.
Edit
If you're after a queue, look into Circular Queues. That's typically how you'd implement one in an array. Alternatively, rather than using an array, try a Linked List - that lets it be arbitrarily big (the only limit is your computer's memory).
You don't need any pointers to shift an array. Just use simple for statement:
int *a; // Your array
int count; // Elements count in array
int length; // Length of array (maxSize)
bool pushFront(const int nPushFront)
{
if (count == length) return false;
for (int i = count - 1; i >= 0; --i)
Swap(a[i], a[i + 1]);
a[0] = nPushFront; ++count;
return true;
}
Without doing your homework for you let me see if I can give you some hints. Implementing a deque (double ended queue) is really quite easy if you can get your head around a few concepts.
Firstly, it is key to note that since we will be popping off the front and/or back in order to efficiently code an algorithm which uses contiguous storage we need to be able to pop front/back without shifting the entire array (what you currently do). A much better and in my mind simpler way is to track the front AND the back of the relevant data within your deque.
As a simple example of the above concept consider a static (cannot grow) deque of size 10:
class Deque
{
public:
Deque()
: front(0)
, count(0) {}
private:
size_t front;
size_t count;
enum {
MAXSIZE = 10
};
int data[MAXSIZE];
};
You can of course implement this and allow it to grow in size etc. But for simplicity I'm leaving all that out. Now to allow a user to add to the deque:
void Deque::push_back(int value)
{
if(count>=MAXSIZE)
throw std::runtime_error("Deque full!");
data[(front+count)%MAXSIZE] = value;
count++;
}
And to pop off the back:
int Deque::pop_back()
{
if(count==0)
throw std::runtime_error("Deque empty! Cannot pop!");
int value = data[(front+(--count))%MAXSIZE];
return value;
}
Now the key thing to observe in the above functions is how we are accessing the data within the array. By modding with MAXSIZE we ensure that we are not accessing out of bounds, and that we are hitting the right value. Also as the value of front changes (due to push_front, pop_front) the modulus operator ensures that wrap around is dealt with appropriately. I'll show you how to do push_front, you can figure out pop_front for yourself:
void Deque::push_front(int value)
{
if(count>=MAXSIZE)
throw std::runtime_error("Deque full!");
// Determine where front should now be.
if (front==0)
front = MAXSIZE-1;
else
--front;
data[front] = value;
++count;
}