Initialization of dynamic memory c++ - c++

In below code, I expect 1 to be initialized in all the 10 elements in x array. But, it doesn't seem to be working. May I know what I am missing here?
int main() {
int *x = new int[10];
for(int i =0; i <10; ++i){
*x = 1;
x++;
}
for(int i = 0; i < 10; ++i)
std::cout<<i<<" is "<<x[i]<<std::endl;
return 0;
}

By the time you end the initialization loop, your x is pointing beyond the last allocated element. Before second forloop you need to readjust x to point to start of memory
x -= 10;
Even better would be to keep your walking pointer as a copy
int *xcopy = x;
for(int i =0; i <10; ++i){
*xcopy = 1;
xcopy++;
}
Or use indexing to update the value
for(int i =0; i <10; ++i){
x[i] = 1;
}

You're incrementing x in the first loop, so after that it points past the end of the array. The second loop uses that new value to read from memory beyond the end, giving undefined behaviour.
The best option is to stop juggling pointers, with the bonus feature of fixing the memory leak:
std::vector<int> x(10, 1);
for(size_t i = 0; i < x.size(); ++i) {
std::cout<<i<<" is "<<x[i]<<std::endl;
}
Otherwise, you could assign to x[i] rather than *x in the first loop and leave x unmodified; or you use a copy of the pointer, or undo the modification after the loop.

Your code has undefined behavior, as you're accessing some random memory (you have changed x).
You should not change x (the pointer). Instead, you can use x[ i ] = 1; or use some other temp pointer to "loop over" x.
You can also decrement x, as #Mohit Jain states, but I wouldn't recommend changing where x points at all.

Related

Memory issue when freeing 2D arrays

2D Array creation phase :
int **table = new int*[10];
for (int a = 0; a < 10; a++)
{
table[a] = new int[10];
for(int b = 0; b < 10; b++){
table[a][b] = 0;
}
}
2D Array deletion phase :
for (int i = 0; i < 10; i++)
{
delete [] table[i];
}
delete [] table;
I noticed something while debugging my code with GDB. All values from table[0][0] to table[0][3] are garbage but all values from table[0][4] to table[0][9] are all 0. This problem is Valid for all pointers from table[0] to table[9].
Then I thought there was a problem with GDB and I printed table[0][0] and table[0][5] to the screen. Indeed the value of table[0][0] is garbage, the value of table[0][5] is 0.
My question is does c++ actually freeing this data?
Yes, the data is actually being freed. If you dereference a pointer after its value is freed, you get undefined behaviour which might be its original value or some other garbage value.

Memory allocation with looping in C++

I have a List class for char arrays. And I want to push back N arrays from 'a' to 'a..a'.
char* x;
SList list;
for (unsigned int i = 1; i < n+1; i++) {
x = new char[i+1];
for (unsigned int j = 0; j < i; j++) {
x[j] = 'a';
}
x[i] = '\0';
list.push_back(&x);
}
But every time, x has the same address. And in result, my List object contains N pointers to the same address.
Is there a way to push back these arrays in loop with correct memory allocation?
Before asking found this question, but it doesn't provide a cool solution for my problem.
In each iteration of the loop x = new char[i+1]; returns a different address, which is stored in x. So the value of x changes in each iteration, yet, the address of x doesn't.
Did you mean
list.push_back(x);
to add the address of the newly allocated memory? Of course this would require you to change the type of list the a collection of char *.
It must be mentioned that dereferencing &x after x goes out of scope will lead to undefined behaviour, because x doesn't exist anymore. You fill list with dangling pointers.
Finally I'd like to mention that you could avoid the nasty manual memory management and simply use a std::vector<std::string>.
std::vector<std::string> list;
for (unsigned int i = 1; i < n+1; i++) {
std::string newString(i, 'a'); // string with i times 'a'
list.push_back(newString);
}
Ok. I found a pretty straightforward solution:
char** x = new char*[n];
SList sl;
for (unsigned int i = 0; i < n; i++) {
x[i] = new char[i+1];
for (unsigned int j = 0; j < i; j++) {
x[i][j] = 'a';
}
x[i][i] = '\0';
sl.push_back(&x[i]);
}
With having N addresses to store pointers to arrays. I can just add their addresses to my list object

The array in my singleton isn't keeping information after leaving a function and then it crashes when it tries to access the information again

I have an array called int **grid that is set up in Amazon::initGrid() and is made to be a [16][16] grid with new. I set every array value to 0 and then set [2][2] to 32. Now when I leave initGrid() and come back in getGrid() it has lost its value and is now 0x0000.
I don't know what to try, the solution seems to be really simple, but I'm just not getting it. Somehow the data isn't being kept in g_amazon but I could post the code.
// Returns a pointer to grid
int** Amazon::getGridVal()
{
char buf[100];
sprintf_s(buf, "Hello %d\n", grid[2][2]);
return grid;
}
int Amazon::initGrid()
{
int** grid = 0;
grid = new int* [16];
for (int i = 0; i < 16; i++)
{
grid[i] = new int[16];
for (int j = 0; j < 16; j++)
{
grid[i][j] = 0;
}
}
grid[2][2] = 32;
return 0;
}
int **grid;
g_amazon = Amazon::getInstance();
g_amazon->initGrid();
grid = g_amazon->getGridVal();
for (int i = 0; i < 16; i++)
{
for (int j = 0; j < 16; j++)
{
int index;
index = (width * 4 * i) + (4 * j);
int gridval;
gridval = grid[i][j];
lpBits[index] = gridval;
lpBits[index + 1] = gridval;
lpBits[index + 2] = gridval;
}
}
It crashes when I run it at the line where sprintf_s prints out [2][2] and it also crashes when I get to gridval = grid[i][j] because it's at memory location 0x000000.
The variable
int** grid
in the initGrid() function is a local variable. Edit** When the function returns the variable is popped off the stack. However, since it was declared with the new operator the memory still exists on the heap; it is simply just not pointed to by your global grid variable.
#Dean said in comment:
I have grid as an int** grid; in class Amazon {}; so shouldn't it stay in memory or do I need a static var.
That is the problem:
local int **grid; on Amazon::initGrid::
is masking
member int **grid; on Amazon::
as the first context has higher priority in name lookup.
So initGrid() allocates memory referenced only by a local pointer. That pointer no longer exists when you return from this function, Amazon::grid was never touched on initialization and you're also left with some bad memory issues.
So, as commented by #Remy-Lebeau, I also suggest
Consider using std::vector> or std::array, 16> instead. There is no good reason to use new[] manually in this situation.

Function behaves badly when passing dynamically allocated pointer

I have this function
void shuffle_array(int* array, const int size){
/* given an array of size size, this is going to randomly
* attribute a number from 0 to size-1 to each of the
* array's elements; the numbers don't repeat */
int i, j, r;
bool in_list;
for(i = 0; i < size; i++){
in_list = 0;
r = mt_lrand() % size; // my RNG function
for(j = 0; j < size; j++)
if(array[j] == r){
in_list = 1;
break;
}
if(!in_list)
array[i] = r;
else
i--;
}
}
When I call this function from
int array[FIXED_SIZE];
shuffle_array(array, FIXED_SIZE);
everything goes all right and I can check the shuffling was according to expected, in a reasonable amount of time -- after all, it's not that big of an array (< 1000 elements).
However, when I call the function from
int *array = new int[dynamic_size];
shuffle_array(array, dynamic_size);
[...]
delete array;
the function loops forever for no apparent reason. I have checked it with debugging tools, and I can't say tell where the failure would be (in part due to my algorithm's reliance on random numbers).
The thing is, it doesn't work... I have tried passing the array as int*& array, I have tried using std::vector<int>&, I have tried to use random_shuffle (but the result for the big project didn't please me).
Why does this behavior happen, and what can I do to solve it?
Your issue is that array is uninitialized in your first example. If you are using Visual Studio debug mode, Each entry in array will be set to all 0xCC (for "created"). This is masking your actual problem (see below).
When you use new int[dynamic_size] the array is initialized to zeros. This then causes your actual bug.
Your actual bug is that you are trying to add a new item only when your array doesn't already contain that item and you are looking through the entire array each time, however if your last element of your array is a valid value already (like 0), your loop will never terminate as it always finds 0 in the array and has already used up all of the other numbers.
To fix this, change your algorithm to only look at the values that you have put in to the array (i.e. up to i).
Change
for(j = 0; j < size; j++)
to
for(j = 0; j < i; j++)
I am going to guess that the problem lies with the way the array is initialized and the line:
r = mt_lrand() % size; // my RNG function
If the dynamically allocated array has been initialized to 0 for some reason, your code will always get stack when filling up the last number of the array.
I can think of the following two ways to overcome that:
You make sure that you initialize array with numbers greater than or equal to size.
int *array = new int[dynamic_size];
for ( int i = 0; i < dynnamic_size; ++i )
array[i] = size;
shuffle_array(array, dynamic_size);
You can allows the random numbers to be between 1 and size instead of between 0 and size-1 in the loop. As a second step, you can subtract 1 from each element of the array.
void shuffle_array(int* array, const int size){
int i, j, r;
bool in_list;
for(i = 0; i < size; i++){
in_list = 0;
// Make r to be betwen 1 and size
r = rand() % size + 1;
for(j = 0; j < size; j++)
if(array[j] == r){
in_list = 1;
break;
}
if(!in_list)
{
array[i] = r;
}
else
i--;
}
// Now decrement the elements of array by 1.
for(i = 0; i < size; i++){
--array[i];
// Debugging output
std::cout << "array[" << i << "] = " << array[i] << std::endl;
}
}
You are mixing C code with C++ memory allocation routines of new and delete. Instead stick to pure C and use malloc/free directly.
int *array = malloc(dynamic_size * sizeof(int));
shuffle_array(array, dynamic_size);
[...]
free(array);
On a side note, if you are allocating an array using the new[] operator in C++, use the equivalent delete[] operator to properly free up the memory. Read more here - http://www.cplusplus.com/reference/new/operator%20new[]/

C++ segmentation fault array

In C++ I get a segmentation fault after telling the program how big the array should be (x).
Why is this happening and how do I fix this?
#include <iostream>
using namespace std;
int main()
{
int x;
cin >> x;
int array[x];
for (int *j=array; j; j++)
{
*j=0;
}
for (int *i=array; i; i++)
{
cin >> *i;
}
cout << array[3] << endl;
}
Your loop conditions are wrong.
for (int *j = array; j; j++)
and
for (int *i=array; i; i++)
will not stop at the end of the array, as the condition j (i) is true when traversing the array (i.e., to be false, the pointer needs to be nullptr). In fact, pointer arithmetic past the array boundary plus one results in undefined behaviour. Your stopping condition should be
i < array + x;
Moreover, variable length arrays are an extension and not support by the C++ standard. Use new[] instead to allocate memory, as #Joshua Byer pointed out.
int * array;
array= new int [x];
http://www.cplusplus.com/doc/tutorial/dynamic/
The conditions used in your loops are incorrect.
eg. for (int *j = array; j; j++) even though j will eventually reach the end of the array but will still never evaluate to false (allowing the loop to finish). On top of this it means you will iterate to past the end of the array and move into Undefined Behaviour, this is probably why you are seeing the segfault.
you either need to do the following (super gross solution!!!! also not C++ standard supported):
for (int i = 0, *j = array; i < x; i++, j++)
which will increment a counter and check the array at the same time as incrementing your pointer.
OR
USE VECTORS
std::vector is a much easier way to do what you are doing.
int arraySize;
cin >> arraySize;
std::vector<int> array(arraySize, 0);
for (int i=0; i < arraySize; i++)
{
cin >> array[i];
}
cout << array.at(3) << endl;
Here is a live example.
Within a for statement, the second expression should terminate the loop by evaluating to false. In this case however you never terminate the loop:
for (int *j=array; j; j++)
Instead do:
for (int *j=array; j < array + x; j++)
The expression array + x (by pointer arithmetic) means one element past the end of the array.
The above goes for both of the loops.