I have int arr[10] and I want the change the array size manually to be 5
I tried:
int arr[10];
for (int i = 0; i < 10; i++)
{
arr[i] = i + 1;
}
arr[5] = nullptr;
But this didn't work and the array size still 10.
Arrays have a fixed size. One of the many reasons you should be using a vector instead. It's easy to change the size of a vector.
Like this
#include <vector>
std::vector<int> arr(10); // vector has size 10, note use (10) not [10]
for (int i = 0; i < arr.size(); i++) // use size() to get the vector's size
{
arr[i] = i + 1;
}
arr.resize(5); // change the vector's size to 5
An array's size is static in C++.
You should use vectors instead.
Please see: How to resize array in C++?
You can't change the size of the array that is not declared dynamically I think.
what you can do is declare a new array of size 5 and copy the contents over to the newly declared array.
Or if you don't want that headache, just use vectors.
Related
I hope to use vector to process the 2d array data obtained by calling a third-party library.
Although I can simply use the loop to assign values one by one, But I prefer to use methods such as insert and copy to deal with this.
I found that reserve doesn't seem to work here. So I used resize instead.
double **a = new double *[1024];
for (int i = 0; i < 1024; ++i) {
a[i] = new double[512];
}
std::vector<std::vector<double>> a_v;
a_v.resize(1024, std::vector<double>(512));
// Copy a -> a_v
I made these attempts:
// Not Working, just 0 in vector
for (int i = 0; i < 1024; ++i){
a_v[i].insert(a_v[i].end(), a[i], a[i] + 512);
}
Is there any good way to solve this problem.
For a 1D array I write like this:
double *b = new double[1024];
std::vector<double> b_v;
b_v.reserve(1024);
b_v.insert(b_v.end(), b, b + 1024);
If the size of the source array is fixed, it is strongly recommended to use std::array instead of std::vector. std::array has continuous memory layout for multidimensional structures, thus std::memcpy can be used for copy if the source array is also continuous in memory.
Look back to the original question. If you want to construct a std::vector<std::vector<double>> from the source array, use a single loop to construct 1D vectors from the source:
std::vector<std::vector<double>> a_v;
a_v.reserve(1024);
for (int i = 0; i < 1024; ++i) {
a_v.emplace_back(std::vector<double>(&(a[i][0]), &(a[i][512])));
}
If there is already a std::vector<std::vector<double>> with the proper size, and you literally just want to do a copy from the source, use the assign member function:
for (int i = 0; i < 1024; ++i) {
a_v[i].assign(&(a[i][0]), &(a[i][512]));
}
std::vector<std::vector<double>> a_v;
a_v.resize(1024, std::vector<double>(512));
is just
std::vector<std::vector<double>> a_v{1024, std::vector<double>(512)};
Unfortunately there is no vector constructor that takes over ownership of a C-style array. So you have to copy all 1024 * 512 doubles. And with the above definition of the vector you needlessly initialize all the doubles before you overwrite them.
You can do it with reserve so none of the double get initialized before you overwrite them and no vector gets copied or moved:
std::vector<std::vector<double>> a_v;
a_v.reserve(1024);
for (std::size_t i = 0; i < 1024; ++i) {
a_v.emplace_back();
std::vector<double> &b_v = a_v.back();
b_v.reserve(512);
b_v.insert(b_v.end(), a[i], a[i] + 512);
}
The following code is used to demonstrate how to insert a new value in a dynamic array:
#include <iostream>
int main()
{
int* items = new int[5] {1, 2, 3, 4, 5}; // I have 5 items
for (int i = 0; i < 5; i++)
std::cout << items[i] << std::endl;
// oh, I found a new item. I'm going to add it to my collection.
// I do this by
// (1) allocating a bigger dynamic array
// (2) copying the existing elements from the old array to the new array
// (3) deleting the old array, redirecting its pointer to the new array
int* items_temp = new int[6];
for (int i = 0; i < 5; i++)
items_temp[i] = items[i];
items_temp[5] = 42;
delete[] items;
items = items_temp;
for (int i = 0; i < 6; i++)
std::cout << items[i] << std::endl;
delete[] items;
}
I am confused about the necessity of using it over a regular array. Can't I just do the same thing with a regular array? Basically, you just define a new array with a larger size and move elements in the previous array to this new array. Why is it better to use a dynamic array here?
You are right, the example you are looking at isn't very good at demonstrating the need for dynamic arrays, but what if instead of going from size 5->6, we had no idea how many items we found until we need to add until the code is actually running?
Regular arrays need to be constructed with their size known at compile time
int foo [5] = { 16, 2, 77, 40, 12071 };
But dynamic arrays can be be assigned as size at runtime
int* Arrary(int size) {
return new int[size];
}
So if you don't know the size of your array, or it may need to grow/shrink you need to use a dynamic array (or better yet just use a std::vector).
Suppose you want to do what you mentioned a multiple times.
for(int i = 0; i < some_val; ++i)
{
int val_to_add;
std::cin >> val_to_add;
int* new_arr = new int[old_size + 1]; // the value is suppsoed to be big in order to indicate that it takes much memory.
copy_old_to_new(new_arr, old_arr); //some function which does the copying.
new_arr[old_size + 1] = val_to_add;
delete[] old_arr;
old_arr = new_arr;
}
Now think about what would happen if we tried to do the same with static arrays.
We wouldn't be able to remove the memory allocated by the old_arr, and the program would use a lot of memory.
We wouldn't be able to construct an array which would be accessible outside the loop, which, obviously, is not intended.
In your example it is not much clear how the usage of dynamic arrays would make use in your intention. So if you feel you could do the same without dynamic arrays, do it without them.
int higher_element = arr[0];
for(int i = 0; i < length; i++)
if(arr[i] > higher_element)
higher_element = arr[i];
cout << "Higher element in an unsorted array :" << higher_element << endl;
int Hash[higher_element] = {0};
Here I want to create a new array of size higher_element and initialize it to 0 but array is not creating, only a garbage value is created.
The output of the higher element is 12.
Since you are using C++, I suggest you to use vector.
Here's the std::vector solution for your problem.
std::vector<int> Hash(higher_element);
Vectors initialize to 0 automatically. But for your clarification,
std::vector<int> Hash(higher_element,0);
You can only use const in declaring the array.
If you want to use a variable to define the size of the array, try this
int *Hash;
Hash = new int[higher_element];
Hope to help you.
I have this function:
void reverse(int* nums, unsigned int size)
This function is supposed to reverse the values in the array it is getting.
Now for reversing I thought to create another array with the size of the array passed in. Assigning this new one from the end of the original array to the start.
But I am a kind of new in C++, So I don't know how to create dynamic array in the size of the parameter of the function.
It's actually not necessary to allocate a new array here. See if you can find a way to solve this problem just by rearranging the existing elements in-place.
Given that this seems like it's an exercise with pointers, you can allocate space by using the new[] operator:
int* auxiliaryArray = new int[size];
You'd then free it by writing
delete[] auxiliaryArray;
However, this isn't the preferred way of doing this in C++. The better route is to use std::vector, which does all its own memory management. That would look like this:
std::vector<int> auxSpace(size);
You can then access elements using the square brackets as you could in a real array. To do this, you'll need to #include <vector> at the top of your program.
In C++, the recommended way to create an array of variable size would be to use an std::vector
#include <vector>
void reverse(int* nums, unsigned int size)
{
std::vector<int> V(size);
...
}
But that approach isn't the best here for performance because it requires additional memory to be allocated of the size of the array, which could be big. It would be better to start from the outside of the array and swap members one by one that are at mirroring positions (so if the size is 5, swap 0 and 4, then swap 1 and 3 and leave 2 alone). This only requires temporary storage of a single int.
You can do it without the need to create another array:
void reverse(int* array, const int size){
for(int i = 0; i < size / 2; i++){
int tmp = array[i];
array[i] = array[size - 1 - i];
array[size - 1 - i] = tmp;
}
}
int main(){
int array[] = {1, 3, 5, 7, 9, 11};
const int size = sizeof(array) / sizeof(array[0]);
reverse(array, size);
for(int i(0); i < size; i++)
std::cout << array[i] << ", ";
}
As you can see above in the loop you only need to swap the first element (element 0) with the n-1 element and the second one with n-1-1 and son on...
Remember arrays are indexed from 0 through n-1.
If you want to allocate new array which is not practical:
int* reverse2(int* array, const int size){
int* tmp = new int[size];
for(int i(size - 1), j(0); j < size; j++, i--)
tmp[j] = array[i];
return tmp;
}
int main(){
int array[] = {1, 3, 5, 7, 9, 11};
for(int i(0); i < size; i++)
std::cout << array[i] << ", ";
std::cout << std::endl;
int* newArray = reverse2(array, size);
for(int i(0) ; i < size; i++)
std::cout << newArray[i] << ", ";
std::cout << std::endl;
delete[] newArray;
return 0;
}
If you want to use a new array you can, but I think is to kill flies with a cannon.
Looks like you are using plain C code and not C++. I say that because of the signature of the function. The signature of the function in a common C++ code could be something like this other:
void reverse(std::vector& items);
You can reverse the current array without a new array, using the current one. You are passing the pointer to the first item of the array, and the content is not constant so that you can modify it. A better signature for the function could be:
void reverse(int* const nums, const unsigned int size);
Looks like a pointer problem. Think about the boundaries to iterate the positions of the array. Would you need to iterate the whole array? Maybe only half array? ;)
As bonus track, what about to exchange the values without an auxiliar variable? (this is true into this case that we are using the fundamental type int... remember the binary arithmetic).
array[pos_head] ^= array[pos_tail];
array[pos_tail] ^= array[pos_head];
array[pos_head] ^= array[pos_tail];
I need to create a square matrix of a given size. I know how to create a dynamic one-dimensional array of a given size. Doesn't the same work for two dimensinal arrays like the lines below?
cin>>size;
int* a[][]=new int[size][size]
int* a[][]=new int[size][size]
No, this doesn't work.
main.cpp:4: error: only the first dimension of an allocated array may have dynamic size
new int[size][size];
^~~~
If the size of the rows were fixed then you could do:
// allocate an array with `size` rows and 10 columns
int (*array)[10] = new int[size][10];
In C++ you can't have raw arrays with two dimensions where both dimensions are dynamic. This is because raw array indexing works in terms of pointers; for example, in order to access the second row a pointer to the first needs to be incremented by the size of the row. But when the size of a row is dynamic the array doesn't know that size and so C++ doesn't know how to figure out how to do the pointer increment.
If you want an array with multiple dynamic dimensions, then you need to either structure the array allocations such that C++'s default array indexing logic can handle it (such as the top answers to this duplicate question), or you need to implement the logic for figuring out the appropriate pointer increments yourself.
For an array where each row has the same size I would recommend against using multiple allocations such as those answers suggest, or using a vector of vectors. Using a vector of vectors addresses the difficulty and dangerousness of doing the allocations by hand, but it still uses more memory than necessary and doesn't allow faster memory access patterns.
A different approach, flattening the multi-dimensional array, can make for code as easy to read and write as any other approach, doesn't use extra memory, and can perform much, much better.
A flattened array means you use just a single dimentional array that has the same number of elements as your desired 2D array, and you perform arithmetic for converting between the multi-dimensional indices and the corresponding single dimensional index. With new it looks like:
int *arr = new int[row_count * column_count];
Row i, column j in the 2d array corresponds to arr[column_count*i + j]. arr[n] corresponds to the element at row n/column_count and column n% column_count. For example, in an array with 10 columns, row 0 column 0 corresponds to arr[0]; row 0, column 1 correponds to arr[1]; row 1 column 0 correponds to arr[10]; row 1, column 1 corresponds to arr[11].
You should avoid doing manual memory management using raw new and delete, such as in the case of int *arr = new int[size];. Instead resource management should be wrapped up inside a RAII class. One example of a RAII class for managing dynamically allocated memory is std::vector.
std::vector<int> arr(row_count * column_count);
arr[column_count*i + j]
You can further wrap the logic for computing indices up in another class:
#include <vector>
class Array2d {
std::vector<int> arr;
int columns;
public:
Array2d(int rows, int columns)
: arr(rows * columns)
, columns(columns)
{}
struct Array2dindex { int row; int column; };
int &operator[] (Array2dindex i) {
return arr[columns*i.row + i.column];
}
};
#include <iostream>
int main() {
int size;
std::cin >> size;
Array2d arr(size, size);
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
arr[{i, j}] = 100;
}
}
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
std::cout << arr[{i, j}] << ' ';
}
std::cout << '\n';
}
}
If you're using C++11 you can also use std::array.
const int iRows = 3, iCols = 3; // number of rows and columns
std::array<std::array<int, iCols>, iRows> matrix;
// fill with 1,2,3 4,5,6 7,8,9
for(int i=0;i<iRows;++i)
for(int j=0;j<iCols;++j)
matrix[i][j] = i * iCols + j + 1;
This class also allows for bounds checking by using the function
std::array::at
which (just like operator[]) returns a const reference if the array-object is const-qualified or a reference if it is not. Please note that
std::array
is not a variable-sized array-type, like
std::vector
You can use std::vector:
std::vector<std::vector<int*>> a(size, std::vector<int*>(size));
This will create a dynamically allocated 2D array of int* with width and height equal to size.
Or the same with new:
int*** a = new int**[size];
for (size_t i = 0; i < size; ++i)
a[i] = new int*[size];
...
for (size_t i = 0; i < size; ++i)
delete a[i];
delete a;
Note that there's no new[][] operator in C++, you just have to call new[] twice.
However, if you want to do it with new and delete instead of std::vector, you should use smart pointers instead of raw pointers, for example:
std::unique_ptr<std::unique_ptr<int*>[]> a(new std::unique_ptr<int*>[size]);
for (size_t i = 0; i < size; ++i)
a[i].reset(new int*[size]);
...
// No need to call `delete`, std::unique_ptr does it automatically.