I am attempting to implement a buddy memory system using c++, but I am running into some issues. I assign a memory address to an element in an array, and the memory address that is stored in main is different than the one returned by the function. I believe there is an error when the address is stored, resetting it to a "valid" position, but i'm not sure which one is the "correct" address. Code and input are posted below.
void buddy_manager(memoryNode* root, process* processes, int num){
int i = 0, count = 0, removed = 0, j = 0;
process running[num];
while(removed < 5){
if(i % 50 == 0 && count<5){
cout << "Adding process: " << count << endl;
running[count] = processes[count];
running[count].space = my_malloc(root, running[count].memory * 1000);
printf("Made new node # %p \n", running[count].space);
count++;
}
//Check to remove the process from the list omitted
}
char* my_malloc (memoryNode* root, int size) {
//If there are child nodes and the current node is not occupied, visit them!
if(root->left != NULL && root->right != NULL && root->occupied == 0){
if(my_malloc(root->left, size) == NULL){
my_malloc(root->right, size);
}
} else{ //we are at a leaf node
if(root->max_size >= size && root->max_size/2 < size && root->left == NULL && root->right == NULL && root->occupied == 0){ //if the node is good for insertion{
printf("address of new node: %p with slize: %d and element size: %d \n", root->start, root->max_size, size);
root->occupied = 1;
return root->start;
} else if(!(root->max_size > size && root->max_size/2 < size) && root->max_size > 0 && root->occupied == 0 && root->left == NULL && root->right == NULL) { //Grow the tree
root->left = new memoryNode(root->start, root->max_size/2, NULL, NULL, 0);
root->right = new memoryNode((char*)(root->start + (root->max_size)/2), root->max_size/2, NULL, NULL, 0);
//printf("Made new nodes at %p and %p \n", root->left->start, root->right->start);
my_malloc(root->left, size);
} else{
return NULL;
}
}
}
My output is as follows:
address of new node: 0xb6d86008 with size: 156250 and element size: 94000
Made new node # 0xb6d86008
Adding process: 1
address of new node: 0xb6dd24bc with size: 312500 and element size: 193000
Made new node # 0xb6dd2400
Adding process: 2
address of new node: 0xb6e1e970 with size: 156250 and element size: 106000
Made new node # 0xb6e1e900
Adding process: 3
address of new node: 0xb6dc8c25 with size: 39062 and element size: 26000
Made new node # 0xb6dc8c00
Adding process: 4
address of new node: 0xb6e44bca with size: 19531 and element size: 18000
Made new node # 0xb6e44b00
A you can see, the first node is fine, but the rest of the nodes end up getting reset to 00 for their last two digits.
A wild guess, if you're using a 32-bit system:
The size of your process structure is not a multiple of 4 bytes. So the space field in the second element of your process running[num] array is not properly aligned to a 4-byte address.
Adding 4-sizeof(process)%4 bytes at the end of this structure should solve the problem.
A wild guess, if you're using a 64-bit system:
The size of your process structure is not a multiple of 8 bytes. So the space field in the second element of your process running[num] array is not properly aligned to an 8-byte address.
Adding 8-sizeof(process)%8 bytes at the end of this structure should solve the problem.
UPDATE:
If you have verified that the issue above is not the reason behind the problem, then print root->start before returning it from function my_malloc, and make sure that it's always aligned to 4 (or 8) bytes.
I've also noticed that function my_malloc doesn't return a value in all cases, which might be leading to undefined behavior (I don't understand how you managed to compile it without an error to begin with).
So inside this function, add return before my_malloc(root->left, size).
Related
I am currently trying to formulate a mergeSort mechanism for a singly linked list. Through research and finding consistent ideas about A) a merge sort being the best way to sort a singly linked list, and B) that these are the key components for performing such an operation, I have arrived at this following code. It almost works exactly as intended, but will only return all of the integers larger than the last inputted number. For example, inputting 7, 6, 5, 4, 3, 2, 1 will return 1, 2, 3, 4, 5, 6, 7, but inputting 1, 2, 3, 4, 5 will only return 5. I've used random input orders so it's not a problem localised to just inputting the numbers in reverse order, but literally any order. If a number is smaller than the final number, it gets removed from the list in the sort process. I cannot locate the cause for this at all. My original problem was caused by an errant while loop that was stopping the iterations after one go, so once I removed that the merge sort was working, but for this problem I have just described.
Any and all advice or suggestions are more than welcome, and thanks for any input you have. My knowledge of linked lists and recursion isn't the greatest, so I really welcome all input/constructive criticism here.
public Node mergeSort(Node head) {
if (head == null || head.getNext() == null) return head;
Node midpoint = findMidpoint(head);
Node rightliststart = midpoint.getNext();
midpoint.setNext(null);
Node rightlist = mergeSort(rightliststart);
Node sorted = sort(leftlist, rightlist);
return sorted;
}
public Node findMidpoint(Node head) {
if (head == null) return head;
Node slowpointer = head;
Node fastpointer = slowpointer.getNext();
while (fastpointer != null) {
fastpointer = fastpointer.getNext();
if (fastpointer != null) {
slowpointer = slowpointer.getNext();
fastpointer = fastpointer.getNext();
}
}
return slowpointer;
}
public Node sort(Node one, Node two) {
Node temp = null;
if (one == null) return two;
if (two == null) return one;
if (one.getData() <= two.getData()) {
temp = one;
temp.setNext(sort(one.getNext(), two));
} else {
temp = two;
temp.setNext(sort(one, two.getNext()));
}
return temp;
}
Example merge code. This shows how the dummy node is used to simplify the code (avoids special case to update head on first node merged).
// merge two already sorted lists
static Node merge(Node list0, Node list1) {
if(list0 == null)
return list1;
if(list1 == null)
return list0;
Node temp = new Node(); // dummy node
Node dest = temp;
while(true){
if(list0.data <= list1.data){
dest.next = list0;
dest = list0;
list0 = list0.next;
if(list0 == null){
dest.next = list1;
break;
}
} else {
dest.next = list1;
dest = list1;
list1 = list1.next;
if(list1 == null){
dest.next = list0;
break;
}
}
}
return temp.next;
}
Example top down merge sort code. It scans the list one time to get the size of the list to avoid double scanning (fast, slow), only scanning n/2 nodes for each recursive split.
// return size of list
static int size(Node head) {
int i = 0;
while(head != null){
head = head.next;
i++;
}
return i;
}
// advance to node n
static Node advance(Node head, int n) {
while(0 < n--)
head = head.next;
return head;
}
// top down merge sort for single link list entry function
static Node sorttd(Node head) {
int n = size(head);
if(n < 2)
return head;
head = sorttdr(head, n);
return head;
}
// top down merge sort for single link list recursive function
static Node sorttdr(Node head, int n) {
if(n < 2)
return head;
int n2 = (n/2);
Node node = advance(head, n2-1);
Node next = node.next;
node.next = null;
head = sorttdr(head, n2);
next = sorttdr(next, n-n2);
head = merge(head, next);
return head;
}
Example bottom up merge sort code. It uses a small (32) array of lists, where array[i] is a list with 0 (empty slot) or 2^i nodes. array[{0 1 2 3 4 ...}] = sorted sub-lists with 0 or {1 2 4 8 16 ...} nodes. Nodes are merged into the array one at a time. A working list is created via a sequence of merge steps with a caller's list node and the leading non-empty slots in the array. The size of the working list doubles with each merge step. After each non-empty slot is used to merge into the working list, that slot is set to empty. After each sequence of merge steps is done, the first empty slot after the leading non-empty slots is set to the working list. A prior slots will now be empty. Once all nodes are merged into the array, the array is merged into a single sorted list. On a large list that doesn't fit in cache, and with randomly scattered nodes, there will be a lot of cache misses for each node accessed, in which case bottom up merge sort is about 30% faster than top down.
// bottom up merge sort for single link list
static Node sortbu(Node head) {
final int NUMLIST = 32;
Node[] alist = new Node[NUMLIST];
Node node;
Node next;
int i;
// if < 2 nodes, return
if(head == null || head.next == null)
return null;
node = head;
// merge node into array
while(node != null){
next = node.next;
node.next = null;
for(i = 0; (i < NUMLIST) && (alist[i] != null); i++){
node = merge(alist[i], node);
alist[i] = null;
}
if(i == NUMLIST) // don't go past end of array
i--;
alist[i] = node;
node = next;
}
// node == null
// merge array into single list
for(i = 0; i < NUMLIST; i++)
node = merge(alist[i], node);
return node;
}
I'm trying to get the size of a key-value or just a key or just a value in redis.
Using debug object key command returns the serialized size of a key-value if it were to be written to disk and not the actual amount of bytes it is using in memory.
This can be confirmed by checking the source code (based from this Redis: Show database size/size for keys)
https://github.com/antirez/redis/blob/4082c38a60eedd524c78ef48c1b241105f4ddc50/src/debug.c#L337-L343
https://github.com/antirez/redis/blob/4082c38a60eedd524c78ef48c1b241105f4ddc50/src/rdb.c#L663-L671
looking at the source code:
/* Save a string object as [len][data] on disk. If the object is a string
* representation of an integer value we try to save it in a special form */
ssize_t rdbSaveRawString(rio *rdb, unsigned char *s, size_t len) {
int enclen;
ssize_t n, nwritten = 0;
/* Try integer encoding */
if (len <= 11) {
unsigned char buf[5];
if ((enclen = rdbTryIntegerEncoding((char*)s,len,buf)) > 0) {
if (rdbWriteRaw(rdb,buf,enclen) == -1) return -1;
return enclen;
}
}
/* Try LZF compression - under 20 bytes it's unable to compress even
* aaaaaaaaaaaaaaaaaa so skip it */
if (server.rdb_compression && len > 20) {
n = rdbSaveLzfStringObject(rdb,s,len);
if (n == -1) return -1;
if (n > 0) return n;
/* Return value of 0 means data can't be compressed, save the old way */
}
/* Store verbatim */
if ((n = rdbSaveLen(rdb,len)) == -1) return -1;
nwritten += n;
if (len > 0) {
if (rdbWriteRaw(rdb,s,len) == -1) return -1;
nwritten += len;
}
return nwritten;
}
And confirm through redis-cli:
127.0.0.1:6379> set a aaaaaaaaaaaaaaaaaaa
OK
127.0.0.1:6379> debug object a
Value at:0x7f985822f168 refcount:1 encoding:embstr serializedlength:20 lru:11611136 lru_seconds_idle:2
127.0.0.1:6379> set a aaaaaaaaaaaaaaaaaaaaaaaaaaaaa
OK
127.0.0.1:6379> debug object a
Value at:0x7f985827c428 refcount:1 encoding:embstr serializedlength:12 lru:11611147 lru_seconds_idle:1
All the different CLI tools are reporting the serialized size of the object and not the memory size which is the interesting and important one.
As of Redis v4, the MEMORY USAGE command does a much better job at guessing the footprint of a key and its value.
I wanted to ask 2 questions..there is a game and someone created some kind of anti-cheat for it. He also put in a scripting feature, where you can read maximum 4 bytes of a player's memory-adress you choose (from the game's process)..
I paste an examplescript here on how to USE it:
public OnPlayerStateChange(playerid, newstate, oldstate)
{
if(newstate == PLAYER_STATE_DRIVER || newstate == PLAYER_STATE_PASSENGER)
{
// read 1 byte from 0x8CB7A5 representing the radio station player is currently listening to.
CAC_ReadMemory(playerid, 0x8CB7A5, 1);
}
return 1;
}
public CAC_OnMemoryRead(player_id, address, size, const content[])
{
if(address == 0x8CB7A5 && size == 1)
{
// "content" is an array with "size" elements, each cell containing 1 byte.
// We read only 1 byte, so the station id is at index 0.
if(content[0] == 0x07) // 0x07 = Radio X
{
SendClientMessage(player_id, -1, "Radio X is indeed the best station.");
}
else
{
SendClientMessage(player_id, -1, "What the hell are you listening to ?!");
}
}
return 1;
}
So the problem is that I have no idea how I find out a specific address..my goal is to find out an adress of a SPECIFIC .dll (which people inject to "cheat") that I can BAN the person from the game with a script...so I have this specific file on my deskop and I have no idea how to even start to find out what kind of memory-address this file would use if injected..
Thanks for any help in advance :)
I am trying to find the middle element of a double linked list in constant time complexity .
I came across the following http://www.geeksforgeeks.org/design-a-stack-with-find-middle-operation/ solution.
But I don't understand how to use the middle pointer.
Can anyone please help me understand this or give me a better solution .
I've re-written this code in C++ for explanation purposes:
#include <iostream>
typedef class Node* PNode;
class Node{
public:
PNode next;
PNode prev;
int data;
Node(){
next = nullptr;
prev = nullptr;
data = 0;
}
};
class List{
private:
//Attributes
PNode head;
PNode mid;
int count;
//Methods
void UpdateMiddle( bool _add );
public:
//Constructors
List(){
head = nullptr;
mid = nullptr;
count = 0;
}
~List(){
while( head != nullptr ){
this->delmiddle();
std::cout << count << std::endl;
}
}
//Methods
void push( int _data );
void pop();
int findmiddle();
void delmiddle();
};
void List::UpdateMiddle( bool _add ){
if( count == 0 ){
mid = nullptr;
}
else if( count == 1 ){
mid = head;
}
else{
int remainder = count%2;
if(_add){
if( remainder == 0 ){
mid = mid->prev;
}
}
else{
if( remainder == 1 ){
mid = mid->next;
}
}
}
}
void List::push( int _data ){
PNode new_node = new Node();
new_node->data = _data;
new_node->prev = nullptr;
new_node->next = head;
if( head != nullptr ) head->prev = new_node;
head = new_node;
count++;
UpdateMiddle( true );
}
void List::pop(){
if( head != nullptr ){
PNode del_node = head;
head = head->next;
if( head != nullptr ) head->prev = nullptr;
delete del_node;
count--;
UpdateMiddle(false);
}
else if( count != 0 ){
std::cout << "ERROR";
return;
}
}
int List::findmiddle(){
if( count > 0 ) return mid->data;
else return -1;
}
void List::delmiddle(){
if( mid != nullptr ){
if( count == 1 || count == 2){
this->pop();
}
else{
PNode del_mid = mid;
int remainder = count%2;
if( remainder == 0 ){
mid = del_mid->next;
mid->prev = del_mid->prev;
del_mid->prev->next = mid;
delete del_mid;
count--;
}
else{
mid = del_mid->prev;
mid->next = del_mid->next;
del_mid->next->prev = mid;
delete del_mid;
count--;
}
}
}
}
The push and pop functions are self-explanatory, they add nodes on top of the stack and delete the node on the top. In this code, the function UpdateMiddle is in charge of managing the mid pointer whenever a node is added or deleted. Its parameter _add tells it whether a node has been added or deleted. This info is important when there is more than two nodes.
Note that when the UpdateMiddle is called within push or pop, the counter has already been increased or decreased respectively. Let's start with the base case, where there is 0 nodes. mid will simply be a nullptr. When there is one node, mid will be that one node.
Now let's take the list of numbers "5,4,3,2,1". Currently the mid is 3 and count, the amount of nodes, is 5 an odd number. Let's add a 6. It will now be "6,5,4,3,2,1" and count will now be 6 an even number. The mid should also now be 4, as it is the first in the middle, but it still hasn't updated. However, now if we add 7 it will be "7,6,5,4,3,2,1", the count will be 7, an odd number, but notice that the mid wont change, it should still be 4.
A pattern can be observed from this. When adding a node, and count changes from even to odd, the mid stays the same, but from odd to even mid changes position. More specifically, it moves one position to the left. That is basically what UpdateMiddle does. By checking whether count is currently odd or even after adding or deleting a node, it decides if mid should be repositioned or not. It is also important to tell whether a node is added or deleted because the logic works in reverse to adding when deleting. This is basically the logic that is being applied in the code you linked.
This algorith works because the position of mid should be correct at all times before adding or deleting, and function UpdateMiddle assumes that the only changes were the addition or deletion of a node, and that prior to this addition or deletion the position of mid was correct. However, we make sure of this by making the attributes and our function UpdateMiddle private, and making it modifiable through the public functions.
The trick is that you don't find it via a search, rather you constantly maintain it as a property of the list. In your link, they define a structure that contains the head node, the middle node, and the number of nodes; since the middle node is a property of the structure, you can return it by simply accessing it directly at any time. From there, the trick is to maintain it: so the push and pop functions have to adjust the middle node, which is also shown in the code.
More depth: maintaining the middle node: we know given the count that for an odd number of nodes (say 9), the middle node is "number of nodes divided by 2 rounded up," so 9/2 = 4.5 rounded up = the 5th node. So if you start with a list of 8 nodes, and add a node, the new count is 9, and you'll need to shift the middle node to the "next" node. That is what they are doing when they check if the new count is even.
I am writing code to ingest the IOR file generated by the team responsible for the server and use it to bind my client to their object. Sounds easy, right?
For some reason a bit beyond my grasp (having to do with firewalls, DMZs, etc.), the value for the server inside the IOR file is not something we can use. We have to modify it. However, the IOR string is encoded.
What does Visibroker provide that will let me decode the IOR string, change one or more values, then re-encode it and continue on as normal?
I've already looked into IORInterceptors and URL Naming but I don't think either will do the trick.
Thanks in advance!
When you feel like you need to hack an IOR, resist the urge to do so by writing code and whatnot to mangle it to your liking. IORs are meant to be created and dictated by the server that contains the referenced objects, so the moment you start mucking around in there, you're kinda "voiding your warranty".
Instead, spend your time finding the right way to make the IOR usable in your environment by having the server use an alternative hostname when it generates them. Most ORBs offer such a feature. I don't know Visibroker's particular configuration options at all, but a quick Google search revealed this page that shows a promising value:
vbroker.se.iiop_ts.host
Specifies the host name used by this server engine.
The default value, null, means use the host name from the system.
Hope that helps.
Long time ago I wrote IorParser for GNU Classpath, the code is available. It is a normal parser written being aware about the format, should not "void a warranty" I think. IOR contains multiple tagged profiles that are encapsulated very much like XML so we could parse/modify profiles that we need and understand and leave the rest untouched.
The profile we need to parse is TAG_INTERNET_IOP. It contains version number, host, port and object key. Code that reads and writes this profile can be found in gnu.IOR class. I am sorry this is part of the system library and not a nice piece of code to copy paste here but it should not be very difficult to rip it out with a couple of dependent classes.
This question has been repeatedly asked as CORBA :: Get the client ORB address and port with use of IIOP
Use the FixIOR tool (binary) from jacORB to patch the address and port of an IOR. Download the binary (unzip it) and run:
fixior <new-address> <new-port> <ior-file>
The tool will override the content of the IOR file with the 'patched' IOR
You can use IOR Parser to check the resulting IOR and compare it to your original IOR
Use this function to change the IOR. pass stringified IOR as first argument.
void hackIOR(const char* str, char* newIOR )
{
size_t s = (str ? strlen(str) : 0);
char temp[1000];
strcpy(newIOR,"IOR:");
const char *p = str;
s = (s-4)/2; // how many octets are there in the string
p += 4;
int i;
for (i=0; i<(int)s; i++) {
int j = i*2;
char v=0;
if (p[j] >= '0' && p[j] <= '9') {
v = ((p[j] - '0') << 4);
}
else if (p[j] >= 'a' && p[j] <= 'f') {
v = ((p[j] - 'a' + 10) << 4);
}
else if (p[j] >= 'A' && p[j] <= 'F') {
v = ((p[j] - 'A' + 10) << 4);
}
else
cout <<"invalid octet"<<endl;
if (p[j+1] >= '0' && p[j+1] <= '9') {
v += (p[j+1] - '0');
}
else if (p[j+1] >= 'a' && p[j+1] <= 'f') {
v += (p[j+1] - 'a' + 10);
}
else if (p[j+1] >= 'A' && p[j+1] <= 'F') {
v += (p[j+1] - 'A' + 10);
}
else
cout <<"invalid octet"<<endl;
temp[i]=v;
}
temp[i] = 0;
// Now temp has decoded IOR string. print it.
// Replace the object ID in temp.
// Encoded it back, with following code.
int temp1,temp2;
int l,k;
for(k = 0, l = 4 ; k < s ; k++)
{
temp1=temp2=temp[k];
temp1 &= 0x0F;
temp2 = temp2 & 0xF0;
temp2 = temp2 >> 4;
if(temp2 >=0 && temp2 <=9)
{
newIOR[l++] = temp2+'0';
}
else if(temp2 >=10 && temp2 <=15)
{
newIOR[l++] = temp2+'A'-10;
}
if(temp1 >=0 && temp1 <=9)
{
newIOR[l++] = temp1+'0';
}
else if(temp1 >=10 && temp1 <=15)
{
newIOR[l++] = temp1+'A'-10;
}
}
newIOR[l] = 0;
//new IOR is present in new variable newIOR.
}
Hope this works for you.