do_something(int array[][])
{
}
int main()
{
int array_length;
cin>> array_length;
int array[array_length][array_length];
for()
{
"putting elements of array"
}
}
I have seen people putting some const int so they can pass array to the function. Question is how do I pass a multidimensional array to the function if I don't know its size until it is entered.
int array[array_length][array_length];
This line will not compile. Unlike some other languages (Java, possibly), allocating an array of dynamic size is different than one of constant size. The [] notation will create a constant-size array, which will not work your attempt to pass it array_length. To allocate this array dynamically, use
int **array = new int*[array_length];
The you'll need to iterate through array and allocate each sub-array to the correct size with
array[i] = new int[array_length];
After this, you'll need to reference array_length when iterating through array, as it is your only indication as to the size of the array.
Related
I am currently working on a program that requires a function to figure out array values and then stores those values in arr1[], which has an unknown size and is calculated within the function. It then passes back the entire array along with the size of the array back to the main function. Using a struct, I returned the size, but I cannot return the array for some reason. Can someone please direct me in the right direction with what I'm doing wrong?
Here is my struct:
struct Arr
{
int size_1;
int arr_1[];
};
And here is part of my function that returns the array, where arr1[] is the array in which I need to return:
Arr smallElement(int arr[], int size)
{
Arr tempArr;
for (int count = 0; count < newSize; count++)
{
tempArr.arr_1[count] = arr1[count];
}
return tempArr;
}
This is what I use in my main function to call the function to print the array, but it just prints 3 random numbers every time:
Arr a;
a = smallElement(array, n);
cout << "The array is: ";
for (int count = 0; count < a.size_1; count++)
{
cout << a.arr_1[count] << " ";
}
Inside a struct, int arr_1[] does not define an actual array. It is actually just a placeholder called a flexible array member. In C, this is used by allocating memory for the fully defined part of the struct plus additional space for the array—you have to manually add space when using malloc or other memory allocation. In standard C++, you should not use this. (It is not part of standard C++. It is an extension adopted by some compilers from C, where it is standard.)
When you declared tempArr, it created just the fully defined portion of the struct. When you assigned values to tempArr.arr_1[count], there was no allocated array there, so the behavior is undefined. When you return tempArr, a copy of the struct is returned, but, again, only a copy of the defined portion.
If the caller knows the size the array will be, you may want to have them pass the array to the function, perhaps by allocating it with new or by using std::vector and passing the vector by reference. If the size is not known by the caller, then you may want to have the function allocate the array and return a pointer to it (in which case the caller is responsible for deleting it later), or you may want to pass an empty (or otherwise available for re-use) std::vector to the function, preferably by reference. Inside the function, it can add more elements to the vector, and the std::vector will manage memory allocation for you.
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 ve got a function that takes 3 parameteres, first one is **double.
normalizeDataZeroMeanUnitSD(double ** trainingActions, int numberOfTrainingActions, int descriptorDimension)
When I call it from main, I am trying to use normalizeDataZeroMeanUnitSD(data, 681, 24); however, I am receiving
cannot convert parameter 1 from 'double [681][24]' to 'double **'
This is how I construct the data array:
fstream infile;
infile.open("gabor\\Data.txt");
double data[681][24];
while (!infile.eof())
{
for(int j=0;j<681;j++)
{
for(int k=0; k<24;k++)
{
infile >> data[j][k];
}
}
}
infile.close();
Is there a way to do the same using **data?
The error is pretty clear: Datatype double [681][24] is not the same as double **. While it's true that double[681] can decay to a pointer to its first element (thus, double*), that does not imply that double[681][24] can decay to double**.
Think about it this way: double** implies a pointer to many pointers. But double[][] does not have ANY pointers in it. At best, an array of ANY dimensions still only has, at very most, one pointer: to the beginning of its contiguous storage.
You could use a template:
template<std::size_t M, std::size_t N>
void normalizeDataZeroMeanUnitSD(double (&trainingActions)[M][N], int descriptorDimension)
{
for( std::size_t m = 0; m < M; ++m )
{
for( std::size_t n = 0; n < N; ++n )
{
trainingActions[m][n] = ...;
}
}
}
But beware of code bloat if you call this with many differently sized arrays.
Use any of the following declarations. Both are equivalent.
NormalizeDataZeroMeanUnitSD(double trainingActions[][24], int numberOfTrainingActions, int descriptorDimension)
NormalizeDataZeroMeanUnitSD(double trainingActions[681][24], int numberOfTrainingActions, int descriptorDimension)
When you declare a 2D array it takes up contiguous memory locations. So you need to specify at least the number of columns (in case of row major architecture).
For row major and column major definitions, have a look at this.
For your edited question, yes you can declare using **data. Dynamically allocate the data array. But remember to free it when you're done with it.
double **data=new double*[681];
for (int i=0;i<681;i++)
{
data[i]=new double[24];
}
//do what you want to do
for (int i=0;i<681;i++)
{
delete [] data[i];
}
delete [] data;
Now your function prototype can be like void func(double **pp) because data is a pointer not a 2D array.
A 2d array is a continuous area of storage. The function expects a pointer to pointers. These are incompatible.
The function expects an array of pointers to arrays; you have an array of arrays. Some options are:
change the function to take a more friendly type, perhaps double* pointing to the first element of a contiguous 2-dimensional array; or
build a separate array of pointers pointing to each row of your 2-dimensional array; or
restructure your data into an array of pointers to arrays.
Here is a constructive answer for how to make it work.
Basically, you need to generate an array that has pointers to each 1D slice of your 2D array.
double data[N][M] = {...};
double *dataPtrs[N];
for(size_t n=0; n<N; ++n) {
dataPtrs[n] = data[n];
}
normalizeDataZeroMeanUnitSD(dataPtrs, N, M); // order of N and M might be wrong
Yay, I get to rant about this again.
In C++, despite having similar syntax, 2D arrays are NOT jagged arrays. 2D arrays (double foo[681][24]) are allocated contiguously in memory. When you deference a 2D array (foo[j][i]) it actually does *(foo+24*i+j). This is all done under the hood. The sizeof(foo)==sizeof(double)*681*24.
Jagged arrays are (double** bar;). This is a bunch of different arrays: first, you allocate an array of pointer, 268 members long. Each pointer will point to an array of doubles, 24 elements long. Bar is just a pointer, so sizeof(bar)==sizeof(void*).
More annoyingly, 2D arrays (or a static array of any dimension) behave the opposite of all other types in C++ in the following reguard: they are passed implicitly by reference, causing the weird phenomenon below.
void foo(double bar[24][24]) { std::cout << sizeof(bar) << std::endl;}
int main() {
double test[24][24];
std::cout << sizeof(test) << std::endl;//returns sizeof(double)*24*24
foo(test);//returns sizeof(void*), implicitly passed by reference, opossite of the rest of c++!
double[][] is not the same thing as double**.
double** is a pointer to pointers.
double[][] is a 2-dimensional array allocated as continuous storage.
In order to pass a "2-dimensional array" to the function, you need to create an array of pointers to arrays. For example:
double* array_2d[681];
for(unsigned int i=0; i<681; ++i) {
array_2d[i] = new double[24];
}
normalizeDataZeroMeanUnitSD(array_2d, 681, 24);
Remember to later delete[] each element of array_2d!
Better yet, change normalizeDataZeroMeanUnitSD to take a reference to std::vector<std::vector<double>>, and you no longer have to worry about memory management, nor passing the correct dimensions to the function.
I'm new to C++ and still really confused about how 2d arrays work with pointers. If I have this (in example format):
int* anarray = anarrayfiller();
for (int a=0;a<10;a++) {
for (int b=0;b<10;b++) {
cout<<(char)anarray[a][b]; //Here's the error mentioned below
}
cout<<"\n";
}
//Later, outside main
int* anarrayfiller() {
int anarray[10][10];
//Populated here
return &anarray;
}
This produces an error under b in the cout<< line:
"Expression must have pointer-to-object type"
I would just check how to search through 2d arrays, and I found this:
A pointer to 2d array
Which suggests that actually this pointer points to the array of ints inside anarray[0], and if that's the case, I must be missing something in terms of returning pointers - wouldn't I then have to return a pointer to a 2d array of pointers that each points to a specific int from anarray? I'm pretty confused here. How do pointers work with 2d arrays?
You have a few errors here:
You return a pointer to a local variable. After the function returns the stack area previously occupied by that variable no longer exist, or is used by the next function.
You return a pointer to an integer, while you have a two-dimensional array. The closest would be a pointer-to-pointer.
You access thing single-pointer as though it was a double-pointer (pointer-to-pointer or pointer-to-array or array-or-arrays), but it's not. That's the reason you get errors at the pointed to line.
But you can't use pointer to pointer, as the memory layout of an array-of-arrays (a two-dimensional array) is different from a pointer-to-pointer. See e.g. this old answer of mine for an explanation of why.
This can be solved most easily by creating the array dynamically on the heap, as a pointer-to-pointer:
int **anarrayfiller()
{
int **anarray = malloc(sizeof(int *) * 10);
for (int i = 0; i < 10; ++i)
{
anarray[i] = malloc(sizeof(int) * 10);
/* Populate here */
}
return anarray;
}
As you tagged your question as C++, you should actually avoid plain arrays or pointers in favor of either std::vector (if you need to add dynamically) or std::array (if you have a fixed compile-time size):
std::array<std::array<int, 10>, 10> anarrayfiller()
{
std::array<std::array<int, 10>, 10> anarray;
// Populate arrays
return anarray;
}
Here is my simple code
arrayfunc() should store some numbers in an array, and return the pointer of this array
to main function where the content of the array would be printed
What is the problem with my code?
It only returns the pointer to the first element of the array
Any help will be appreciated.
Thanks in advance.
#include <iostream>
using namespace std;
//The definition of the function should remain the same
int* arrayfunc()
{
int *array[10];
array[0] =new int;
array[1] =new int;
array[2] =new int;
array[3] =new int;
*array[0]=10;
*array[1]=11;
*array[2]=12;
*array[3]=13;
return *array;
}
int main()
{
for(int i=0;i<4;i++)
cout<<*(arrayfunc()+i)<<endl;
return 0;
}
(1) You should allocate your array with new if you want to return it: int* array = new int[10]; [assuming here you want array of ints and not array of int*'s]
(2) to return the pointer to the first element in the array, use return array and not return *array
(3) your array is array of pointers, and not array of ints.
Your array is allocated on stack, so as soon as the function returns, it's freed. So you want to return a pointer to a dead memory.
But you are not doing that, you are just returning the valid (copy of) value of the 0th array item.
So, what you have to do:
The best idea would be to switch to stl containers. You should be using std::vector or something like that.
If you stick to the idea of manual memory management, you have to allocate the array on heap, return it from the function, and perhaps deallocate it in the caller.
Edit:
basically you want the following:
using namespace std;
vector<int> arrayfunc()
{
vector<int> v;
v.push_back(10);
...
return v;
}
...
vector<int> result = arrayfunc();
cout << result[0] << ...
This would be the right C++ way.
(Nitpicking:) You don't need to care about copying the vector, because of the RVO used by all modern C++ compilers.
Allocating an array on heap should be simple, too:
int* array = new int[4];
array[0] = 10;
...
return array;
...
int* array = arrayfunc();
...
delete[] array;
But I would strongly advise to take the former approach (with vector).
This codes seems wrong to me in several levels.
Never return an internal variable of a function. The variable array is only defined in the function, so it should never be returned outside.
Why do you allocate each int by itself with new? I would allocate the entire array at once. If you know the array length and it's constant, consider having it defined statically.
http://msdn.microsoft.com/en-us/library/s1sb61xd.aspx
Just try return array; instead of return *array;