I am creating a class called intArray. It should behave identical to the std::array but only with integers. I got a bit stuck in the constructor where I need to fill the array with zeros, given the size. I am NOT using the std::array, I am just trying to sort of re-create it.
I was thinking along the lines of having n integers (where n is the fixed array size) that are consecutive in memory and a pointer, head, that points to the first integer... Then the rest of the integers could be easily accessed by getting the i'th memory location after the memory location of the data the head is pointing to.
Basically, how should I initialize/fill an array with zeroes inside its constructor?
Is this a valid way to implement the constructor? If so, is there a better way? If not, how should I accomplish this task?
Sorry, it's a confusing question, just comment if you need any clarity.
Here's my constructor so far:
(not sure how to create 'n' independent integers)
intArray::intArray(size_t n){
intArray::size = n;
int num = 0;
intArray::head = num //head is an int*
//have no idea how to do this...
for(int i=0; i<n; i++){
&num + 1 = 0; //trying to set next value in memory to 0
//obviously not gonna work!
}//end loop
}//end of constructor
I greatly appreciate any answers/comments. Let me now if I missed anything, thanks!
Firstly, what you are really doing is recreating vector, not array, because your construction parameter is not required to be known at compile time. With that in mind, you need to be using dynamic memory unless you change your approach to passing the size as a template parameter and storing an actual array instead of pointer (this is exactly how std::array works).
And so, using dynamic memory, the constructor becomes:
intArray::intArray( size_t n )
: head( new int[ n ] )
, currentSize( n )
{
std::fill( head, head + n, 0 );
}
If you don't want to use std::fill, then the equivalent loop is:
for( size_t i = 0; i < n; i++ )
{
head[i] = 0;
}
Related
I have a dynamically populated array of strings in C++:
string** A;
it is populated like this:
A = new string*[size1];
and then:
for (unsigned int i = 0; i < size1; i++)
{
A[i] = new string[size2];
for (unsigned int j = 0; j < size2; j++)
{
A[i][j] = whatever[j];
}
}
elsewhere, I want to find out the dimensions (size1 and size2).
I tries using this:
sizeof(A[i]) / sizeof(A[i][0])
but it doesn't work.
Any ideas ?
Thanks
When you allocate memory via new T[N], the value N is not stored anywhere . If you need to know it later, you will need to keep track of it in your code.
There are pre-existing classes for allocating memory that also remember the length that was allocated. In your code:
vector<vector<string>> A(size1, vector<string>(size2));
// (code to populate...)
then you can access A.size() to get size1, and A[0].size() to get size2.
If the dimensions are known at compile-time you may use array instead of vector.
It is very simple to find the size of a two dimensional (more exactly of one-dimensional dynamically allocated arrays) array. Just declare it like
std::vector<std::vector<std::string>> A;
and use
std::cout << A.size() << std::endl;
As for your approach then you have to store the sizes in some variables when the array is allocated.
If you are learning C++, I would recommend that you learn Classes. With a class you can encapsulate int variables along with your 2D array that you can use to store the dimensions of your array. For example:
class 2Darray{
string **array;
int rows;
int cols;
}
You can then get the dimensions of your 2Darray object anytime by reading these member variables.
vectors will do this for you behind the scenes but its good for you to learn how to do this.
You can't create an array just using pointer operator. Every array is basically a pointer with allocated memory. That's why compiler wants constant before creating array.
Basically; sizeof(A[i]) won't give you the size of array. Because sizeof() function will return the a pointers size which is points to A[i] location. sizeof(A[i]) / sizeof(A[i][1]) will probably give you 1 because you are basically doing sizeof(int)/sizeof(int*)
So you need to store the boundary yourself or use vectors. I would prefer vectors.
Can't get array dimensions through pointer(s)
I want to have an array accessible by all functions of a class.
I put the array as private variable in the header file.
private:
int* arrayName;
In the .cpp file where I implement the class, the constructor takes in an int value (size) and creates the array. The goal is to fill it up
ClassName::ClassName(int numElements){
arrayName = new int[numElements]; //make arrays the size of numElements
for(int i = 0; i<numElements; i++)
arrayName[i] = 0;
}
I feel like this is quite inefficient. I know you can do int array[5] = {0}; but how do you do it when you don't initially know the size.
If you want to zero-initialize a newed array, just do value-initialize it. This has the effect of zero-initializing its elements:
arrayName = new int[numElements]();
// ^^
But you really want to be using an std::vector<int>.
private:
std::vector<int> vname;
and
ClassName::ClassName(int numElements) : vname(numElements) {}
This way you don't have to worry about deleting an array and implementing copy constructors and assignment operators.
You can use the memset function:
memset(arrayName,0,sizeof(int)*numElements);
This void * memset ( void * ptr, int value, size_t num ); function sets the first num bytes of the block of memory pointed by ptr to the specified value (interpreted as an unsigned char).
To use it you must include the string.h header file.
For more information: http://www.cplusplus.com/reference/cstring/memset/
What you want to do is progressively expand the array on demand.
arrayName = new int[numElements];
for(int i = 0; i<numElements; i++)
arrayName[i] = 0;
The above code (what you gave) will give you an array of size numElements, and THEN the for loop will fill it. This is allocated now, and can't, as I understand it, be simply or easily resized (memset will overwrite previously held values in the array).
You could copy the whole array over every time you want to resize it:
int * oldarr = new int[OldSize];
//fill your old array
int * newarr = new int[NewSize];
for(int i = 0; i<OldSize; i++)
newarr[i] = oldarr[i];
Other than that, you could make the array much larger, or you could use various STLs, such as std::vector. Vector can be increased with a simple push_back function, and allows [] operator access (like arr[5] and whatnot).
Hope this helps!
int (**test)[4][4] = new ???[64];
for (int i = 0; i < 32; ++i)
{
test[i] = new int[4][4][4];
}
I'm trying to create a "list" of pointers that will be initialized to NULL and then later assigned the address of a new multidimensional array of int. The for loop will (eventually) vary in number of iterations, anywhere from 0 to the full 64. I expect to end up with an array of pointers where some are valid and the rest are NULL. The problem is that I can't figure out the syntax for allocating this array of pointers. Basically, what could I put in place of those question marks?
In the name of readability, may I suggest using a typedef?
typedef int (*t)[4][4];
t* test = new t[64];
You will thank me next week when you have to maintain that horrible piece of code ;)
I am attempting to write a template/class that has a few functions, but I'm running into what seems like a rather newbie problem. I have a simple insert function and a display values function, however whenever I attempt to display the value, I always receive what looks like a memory address(but I have no idea), but I would like to receive the value stored (in this particular example, the int 2). I'm not sure how to dereference that to a value, or if I'm just completely messing up. I know that vectors are a better alternative, however I need to use an array in this implementation - and honestly I would like to gain a more thorough understanding of the code and what's going on. Any help as to how to accomplish this task would be greatly appreciated.
Example Output (running the program in the same way every time):
003358C0
001A58C0
007158C0
Code:
#include <iostream>
using namespace std;
template <typename Comparable>
class Collection
{
public: Collection() {
currentSize = 0;
count = 0;
}
Comparable * values;
int currentSize; // internal counter for the number of elements stored
void insert(Comparable value) {
currentSize++;
// temparray below is used as a way to increase the size of the
// values array each time the insert function is called
Comparable * temparray = new Comparable[currentSize];
memcpy(temparray,values,sizeof values);
// Not sure if the commented section below is necessary,
// but either way it doesn't run the way I intended
temparray[currentSize/* * (sizeof Comparable) */] = value;
values = temparray;
}
void displayValues() {
for (int i = 0; i < currentSize; i++) {
cout << values[i] << endl;
}
}
};
int main()
{
Collection<int> test;
int inserter = 2;
test.insert(inserter);
test.displayValues();
cin.get();
return 0;
}
Well, if you insist, you can write and debug your own limited version of std::vector.
First, don't memcpy from an uninitialized pointer. Set values to new Comparable[0] in the constructor.
Second, memcpy the right number of bytes: (currentSize-1)*sizeof(Comparable).
Third, don't memcpy at all. That assumes that Comparable types can all be copied byte-by-byte, which is a severe limitation in C++. Instead:
EDIT: changed uninitialized_copy to copy:
std::copy(values, values + currentSize - 1, temparray);
Fourth, delete the old array when it's no longer in use:
delete [] values;
Fifth, unless the code is going to make very few insertions, expand the array by more than one. std::vector typically increases its size by a factor of 1.5.
Sixth, don't increment currentSize until the size changes. That will change all those currentSize-1s into currentSize, which is much less annoying. <g>
Seventh, an array of size N has indices from 0 to N-1, so the top element of the new array is at currentSize - 1, not currentSize.
Eighth, did I mention, you really should use std::vector.
This line is wrong:
memcpy(temparray,values,sizeof values);
The first time this line is run, the values pointer is uninitialized, so it will cause undefined behavior. Additionally, using sizeof values is wrong since that will always give the size of a pointer.
Another issue:
temparray[currentSize] = value;
This will also cause undefined bahavior because you have only allocated currentSize items in temparray, so you can only access indices 0 to currentSize-1.
There is also an error in your array access.
temparray[currentSize/* * (sizeof Comparable) */] = value;
Remember that arrays start at index zero. So for an array of length 1, you would set temparray[0] = value. Since you increment currentSize at the top of the insert function, you will need to do this instead:
temparray[currentSize-1] = value;
I'm new to programming and I was wondering, how to get the size of an array, that is, get the size of how many elements are inside the array. For example if I declare an array of size 10, but only input 3 elements into the array, how would I determine the size of this array? If I don't know how many elements I placed in initially.
If you declare an array, e.g. int array[10], then its size is always 10 * sizeof(int). There is no way to know how many times you've accessed it; you'd need to keep track of that manually.
You should consider using container classes, e.g. std::vector:
std::vector<int> vec;
vec.push_back(5);
vec.push_back(10);
vec.push_back(42);
std::cout << vec.size() << "\n"; // Prints "3"
If you declare an old-style array of 10 elements, e.g. std::string words[10], the size of the array is always 10 strings. Even with the new style (std::array), it would be a fixed size.
You might be looking for a std::vector<>. This doesn't have a fixed size, but does have a .size() method. Therefore, if you add three elements to it, it will have .size()==3
to get the array size (in number of elements) assuming you do not know it in advance
use sizeof(a)/sizeof(a[0])
see the below example program. I used C but it should carry over to C++ just fine
#include <stdio.h>
int main(){
int a[10];
printf("%d elements\n",sizeof(a)/sizeof(a[0]));
return 0;
}
//output: 10 elements
There's several possible ways, but they depend on your definition.
If you know there is a value the user won't input (also known as a sentinel value), you can use a function like memset, to set the entire array to that unused value. You would then iterate through the list counting all the variables in the list that don't match that value.
The other way is to build your own array class, which counts whenever the array is modified (you'd have to overload the = and [] functions as appropriate).
You could also build a dynamically linked list, so as the user adds variables, the count can either be determined by walking the list or keeping count.
But, if you're taking the array as the basic array, with no idea as to it's actual starting state, and no idea what to expect from the user (given this is your program, this shouldn't occur), then generally speaking, no, there is known way to know this.
You maintain a counter variable count initialized to 0.
Whenever you are adding to array increment the count by 1.
Whenever you are removing from array decrement the count by 1.
anytime count will give you the size of the array.
Suggestion:
int[10] arr;
//init all to null
for (int i =0; i < 10; i++)
arr[i] = 0;
arr[0]=1;
arr[2]=5;
int sz = 0;
for (int j = 0; j < 10; j++)
if (arr[j] != 0) sz++;