Passing array to function with dynamic declaration of array in C/C++ - c++

I want to pass array to function in C/C++ without declaring and assigning it previously.
As in C# we can:
fun(new int[] {1, 1}); //this is a function call for void fun(int[]) type function
I am not very sure about the correctness of the above code. But I had done something like this in C#.
How can we pass array to function using dynamic declaration (on the spot declaration)??
I'd tried it using following way. But no luck
fun(int {1,1});
fun(int[] {1,1});
fun((int[]) {1,1});
can't we do so..??

In C99 and later, this:
foo((int[]){2,4,6,8});
is approximately the same as this:
int x[] = {2,4,6,8};
foo(x);
Prior to C99, this is not possible. C++ does not support this syntax either.

In C++11, you can use an initializer list:
#include <initializer_list>
#include <iostream>
void fun(std::initializer_list<int> numbers)
{
for (int x : numbers)
std::cout << x << ' ';
std::cout << '\n';
}
int main()
{
fun( {2, 3, 5, 7, 11, 13, 17, 19} );
}

This is possible in ANSI C99 and C11, but not in C89.

May I suggest using a variable argument list?
The declaration syntax would be as follows:
void Function(int Count, ...) {
va_list va;
va_start(va, Count);
for(int i = 0; i < Count; ++i) {
int nextVar = va_arg(va, int);
//Do some stuff with the array element
}
va_end(va);
}
This function is called with:
Function(4, 2, 3, 1, 5);
2, 3, 1 and 5 are the 'array elements'.
Alternatively, if you 'need' an array to go through within that function, it's easy to move the variable list to a proper (dynamic) array, by using new and delete.
Within the va-function:
int *Array = new int[Count];
for(int i = 0; i < Count; ++i) {
Array[i] = va_arg(va, int);
}
//Do Stuff with Array
delete[] Array;

Related

The increment of a pointer does not work. It's quite hard for me to understand what's what

I have recently learned some basics of pointers in C++. So I understand that an array's name is just simply a pointer set to the first position of the array by default. And I do not really comprehend why the first code works just fine whereas the second code shows an error on the line: tab = tab+1;
Could you explain to me why it's so ambiguous?
#include <iostream>
int main()
{
int *tab;
tab = new int[5] {0, 1, 2, 3, 4};
for(int i=0; i<5; i++)
{
std::cout << *tab << std::endl;
tab++;
}
delete []tab;
}
#include <iostream>
int main()
{
int tab[5] = {0, 1, 2, 3, 4};
for(int i=0; i<5; i++)
{
std::cout << *tab << std::endl;
tab = tab+1; // ERROR
}
}
The first code "works" because tab is declared as an actual pointer, and you can assign another pointer to it. But the code also fails, because it has undefined behavior due to tab no longer pointing at the original memory returned by new[] when delete[] is called.
The second code does not work because tab is declared as an array, not a pointer. It decays into a pointer when needed, but you can't assign a pointer to an array.
So I understand that an array's name is just simply a pointer set to the first position of it by default.
No. That is wrong. Arrays are not pointers and pointers are not arrays.
Because arrays can decay to the address of its first elements you can write int* tab_ptr = tab; to get a pointer to the first element:
#include <iostream>
int main()
{
int tab[5] = {0, 1, 2, 3, 4};
int* tab_ptr = tab;
for(int i=0; i<5; i++)
{
std::cout << *tab_ptr << std::endl;
tab_ptr = tab_ptr+1;
}
}
And because arrays do decay to the address of its first element when passed to functions you can also do this:
#include <iostream>
void foo(int t[]){
for(int i=0; i<5; i++)
{
std::cout << *t << std::endl;
t = t+1;
}
}
int main()
{
int tab[5] = {0, 1, 2, 3, 4};
foo(tab);
}
When you write tab = tab+1 you get the error:
<source>:23:13: error: incompatible types in assignment of 'int*' to 'int [5]'
23 | tab = tab+1;
| ~~~~^~~~~~~
because the array does decay to an address in tab+1, and tab+1 is a pointer, but you cannot assign the resulting pointer to the array tab.
The first version does not have this problem, because there tab is indeed a pointer to a dynamic array.

How can I solve this problem with arrays in C++?

I am working on the Edabit challenge: Get arithmetic mean of the given array. Now I have code like that:
#include <iostream>
int data;
using namespace std;
int mean(int data);
int main()
{
int data[] = { 1, 2, 3, 4 };
cout << mean(data);
}
int mean(int data)
{
double mean = 0;
for (int i = 0; i < sizeof(data) / sizeof(data[0]); i++)
{
mean += data[i];
}
mean /= sizeof(data) / sizeof(data[0]);
}
and I am stuck. I use Visual Studio 2019 on Windows 7 Professional, and I have underlined 3 characters ( data[i], and 2x data[0]). For this x Visual Studio says expression must have pointer-to-object type (Error E0142) and I have no idea what it means with this. I only know what pointer is.
In Visual studio I added the return statement, but while shortening the code here for publishing I forgot to add it. Otherwise, this wasn't the actual problem. Now I mustn't add it in the question because the comments would be wrong. The comments are related to the upper question, but my real question (for future readers stuck on this problem) is rather:
How to pass array as an argument in the function.
Now, that I am more proficient in C++, I know the terminology and how to state it, and I also know that this isn't so clear to a total beginner: you can't just write f(int[] arr) (you can't pass arrays), but you have to write f(int* arr) (you can pass a pointer that points on that array). At that time I couldn't just search it because I didn't know much of C++ terminology.
Arrays decay into pointers (an int* in this case) when passed as argument to functions. Your mean function only accepts one single int.
When the array decays into a pointer the size information is lost. You can however prevent that by defining a function that accepts arrays of the exact type and size you need. This can be done with templates:
#include <iostream>
template<typename T, size_t N>
double mean(const T (&data)[N]) {
double sum = 0;
for (size_t i = 0; i < N; ++i)
{
sum += data[i];
}
return sum / N;
}
int main()
{
int input[] = { 1, 2, 3, 4, 5, 6, 7 };
std::cout << mean(input) << '\n';
}
If you don't want to use templates (or only accept arrays of a certain size), you need to pass the size information on to the function manually:
#include <iostream>
#include <iterator> // std::size
double mean(const int* data, size_t N)
{
double sum = 0;
for (size_t i = 0; i < N; ++i)
{
sum += data[i];
}
return sum / N;
}
int main()
{
int input[] = { 1, 2, 3, 4, 5, 6, 7 };
std::cout << mean(input, std::size(input)) << '\n';
}
Your mean function, well, is mean.
1. It doesn't return a value; there is no return statement.
2. It uses variable name the same as the function (not a recommended coding style).
3. There is a global variable data that is hidden by a local variable data inside main.
4. You're confusing the compiler and the reader: the global data variable is a single int. The local variable in main is an array of int.
You should have the last line be:
return mean;

Passing a 2D array as Pointer? [duplicate]

I have a function which I want to take, as a parameter, a 2D array of variable size.
So far I have this:
void myFunction(double** myArray){
myArray[x][y] = 5;
etc...
}
And I have declared an array elsewhere in my code:
double anArray[10][10];
However, calling myFunction(anArray) gives me an error.
I do not want to copy the array when I pass it in. Any changes made in myFunction should alter the state of anArray. If I understand correctly, I only want to pass in as an argument a pointer to a 2D array. The function needs to accept arrays of different sizes also. So for example, [10][10] and [5][5]. How can I do this?
There are three ways to pass a 2D array to a function:
The parameter is a 2D array
int array[10][10];
void passFunc(int a[][10])
{
// ...
}
passFunc(array);
The parameter is an array containing pointers
int *array[10];
for(int i = 0; i < 10; i++)
array[i] = new int[10];
void passFunc(int *a[10]) //Array containing pointers
{
// ...
}
passFunc(array);
The parameter is a pointer to a pointer
int **array;
array = new int *[10];
for(int i = 0; i <10; i++)
array[i] = new int[10];
void passFunc(int **a)
{
// ...
}
passFunc(array);
Fixed Size
1. Pass by reference
template <size_t rows, size_t cols>
void process_2d_array_template(int (&array)[rows][cols])
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < cols; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
In C++ passing the array by reference without losing the dimension information is probably the safest, since one needn't worry about the caller passing an incorrect dimension (compiler flags when mismatching). However, this isn't possible with dynamic (freestore) arrays; it works for automatic (usually stack-living) arrays only i.e. the dimensionality should be known at compile time.
2. Pass by pointer
void process_2d_array_pointer(int (*array)[5][10])
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < 5; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < 10; ++j)
std::cout << (*array)[i][j] << '\t';
std::cout << std::endl;
}
}
The C equivalent of the previous method is passing the array by pointer. This should not be confused with passing by the array's decayed pointer type (3), which is the common, popular method, albeit less safe than this one but more flexible. Like (1), use this method when all the dimensions of the array is fixed and known at compile-time. Note that when calling the function the array's address should be passed process_2d_array_pointer(&a) and not the address of the first element by decay process_2d_array_pointer(a).
Variable Size
These are inherited from C but are less safe, the compiler has no way of checking, guaranteeing that the caller is passing the required dimensions. The function only banks on what the caller passes in as the dimension(s). These are more flexible than the above ones since arrays of different lengths can be passed to them invariably.
It is to be remembered that there's no such thing as passing an array directly to a function in C [while in C++ they can be passed as a reference (1)]; (2) is passing a pointer to the array and not the array itself. Always passing an array as-is becomes a pointer-copy operation which is facilitated by array's nature of decaying into a pointer.
3. Pass by (value) a pointer to the decayed type
// int array[][10] is just fancy notation for the same thing
void process_2d_array(int (*array)[10], size_t rows)
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < 10; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
Although int array[][10] is allowed, I'd not recommend it over the above syntax since the above syntax makes it clear that the identifier array is a single pointer to an array of 10 integers, while this syntax looks like it's a 2D array but is the same pointer to an array of 10 integers. Here we know the number of elements in a single row (i.e. the column size, 10 here) but the number of rows is unknown and hence to be passed as an argument. In this case there's some safety since the compiler can flag when a pointer to an array with second dimension not equal to 10 is passed. The first dimension is the varying part and can be omitted. See here for the rationale on why only the first dimension is allowed to be omitted.
4. Pass by pointer to a pointer
// int *array[10] is just fancy notation for the same thing
void process_pointer_2_pointer(int **array, size_t rows, size_t cols)
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < cols; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
Again there's an alternative syntax of int *array[10] which is the same as int **array. In this syntax the [10] is ignored as it decays into a pointer thereby becoming int **array. Perhaps it is just a cue to the caller that the passed array should have at least 10 columns, even then row count is required. In any case the compiler doesn't flag for any length/size violations (it only checks if the type passed is a pointer to pointer), hence requiring both row and column counts as parameter makes sense here.
Note: (4) is the least safest option since it hardly has any type check and the most inconvenient. One cannot legitimately pass a 2D array to this function; C-FAQ condemns the usual workaround of doing int x[5][10]; process_pointer_2_pointer((int**)&x[0][0], 5, 10); as it may potentially lead to undefined behaviour due to array flattening. The right way of passing an array in this method brings us to the inconvenient part i.e. we need an additional (surrogate) array of pointers with each of its element pointing to the respective row of the actual, to-be-passed array; this surrogate is then passed to the function (see below); all this for getting the same job done as the above methods which are more safer, cleaner and perhaps faster.
Here's a driver program to test the above functions:
#include <iostream>
// copy above functions here
int main()
{
int a[5][10] = { { } };
process_2d_array_template(a);
process_2d_array_pointer(&a); // <-- notice the unusual usage of addressof (&) operator on an array
process_2d_array(a, 5);
// works since a's first dimension decays into a pointer thereby becoming int (*)[10]
int *b[5]; // surrogate
for (size_t i = 0; i < 5; ++i)
{
b[i] = a[i];
}
// another popular way to define b: here the 2D arrays dims may be non-const, runtime var
// int **b = new int*[5];
// for (size_t i = 0; i < 5; ++i) b[i] = new int[10];
process_pointer_2_pointer(b, 5, 10);
// process_2d_array(b, 5);
// doesn't work since b's first dimension decays into a pointer thereby becoming int**
}
A modification to shengy's first suggestion, you can use templates to make the function accept a multi-dimensional array variable (instead of storing an array of pointers that have to be managed and deleted):
template <size_t size_x, size_t size_y>
void func(double (&arr)[size_x][size_y])
{
printf("%p\n", &arr);
}
int main()
{
double a1[10][10];
double a2[5][5];
printf("%p\n%p\n\n", &a1, &a2);
func(a1);
func(a2);
return 0;
}
The print statements are there to show that the arrays are getting passed by reference (by displaying the variables' addresses)
Surprised that no one mentioned this yet, but you can simply template on anything 2D supporting [][] semantics.
template <typename TwoD>
void myFunction(TwoD& myArray){
myArray[x][y] = 5;
etc...
}
// call with
double anArray[10][10];
myFunction(anArray);
It works with any 2D "array-like" datastructure, such as std::vector<std::vector<T>>, or a user defined type to maximize code reuse.
You can create a function template like this:
template<int R, int C>
void myFunction(double (&myArray)[R][C])
{
myArray[x][y] = 5;
etc...
}
Then you have both dimension sizes via R and C. A different function will be created for each array size, so if your function is large and you call it with a variety of different array sizes, this may be costly. You could use it as a wrapper over a function like this though:
void myFunction(double * arr, int R, int C)
{
arr[x * C + y] = 5;
etc...
}
It treats the array as one dimensional, and uses arithmetic to figure out the offsets of the indexes. In this case, you would define the template like this:
template<int C, int R>
void myFunction(double (&myArray)[R][C])
{
myFunction(*myArray, R, C);
}
anArray[10][10] is not a pointer to a pointer, it is a contiguous chunk of memory suitable for storing 100 values of type double, which compiler knows how to address because you specified the dimensions. You need to pass it to a function as an array. You can omit the size of the initial dimension, as follows:
void f(double p[][10]) {
}
However, this will not let you pass arrays with the last dimension other than ten.
The best solution in C++ is to use std::vector<std::vector<double> >: it is nearly as efficient, and significantly more convenient.
Here is a vector of vectors matrix example
#include <iostream>
#include <vector>
using namespace std;
typedef vector< vector<int> > Matrix;
void print(Matrix& m)
{
int M=m.size();
int N=m[0].size();
for(int i=0; i<M; i++) {
for(int j=0; j<N; j++)
cout << m[i][j] << " ";
cout << endl;
}
cout << endl;
}
int main()
{
Matrix m = { {1,2,3,4},
{5,6,7,8},
{9,1,2,3} };
print(m);
//To initialize a 3 x 4 matrix with 0:
Matrix n( 3,vector<int>(4,0));
print(n);
return 0;
}
output:
1 2 3 4
5 6 7 8
9 1 2 3
0 0 0 0
0 0 0 0
0 0 0 0
Single dimensional array decays to a pointer pointer pointing to the first element in the array. While a 2D array decays to a pointer pointing to first row. So, the function prototype should be -
void myFunction(double (*myArray) [10]);
I would prefer std::vector over raw arrays.
We can use several ways to pass a 2D array to a function:
Using single pointer we have to typecast the 2D array.
#include<bits/stdc++.h>
using namespace std;
void func(int *arr, int m, int n)
{
for (int i=0; i<m; i++)
{
for (int j=0; j<n; j++)
{
cout<<*((arr+i*n) + j)<<" ";
}
cout<<endl;
}
}
int main()
{
int m = 3, n = 3;
int arr[m][n] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
func((int *)arr, m, n);
return 0;
}
Using double pointer In this way, we also typecast the 2d array
#include<bits/stdc++.h>
using namespace std;
void func(int **arr, int row, int col)
{
for (int i=0; i<row; i++)
{
for(int j=0 ; j<col; j++)
{
cout<<arr[i][j]<<" ";
}
printf("\n");
}
}
int main()
{
int row, colum;
cin>>row>>colum;
int** arr = new int*[row];
for(int i=0; i<row; i++)
{
arr[i] = new int[colum];
}
for(int i=0; i<row; i++)
{
for(int j=0; j<colum; j++)
{
cin>>arr[i][j];
}
}
func(arr, row, colum);
return 0;
}
You can do something like this...
#include<iostream>
using namespace std;
//for changing values in 2D array
void myFunc(double *a,int rows,int cols){
for(int i=0;i<rows;i++){
for(int j=0;j<cols;j++){
*(a+ i*rows + j)+=10.0;
}
}
}
//for printing 2D array,similar to myFunc
void printArray(double *a,int rows,int cols){
cout<<"Printing your array...\n";
for(int i=0;i<rows;i++){
for(int j=0;j<cols;j++){
cout<<*(a+ i*rows + j)<<" ";
}
cout<<"\n";
}
}
int main(){
//declare and initialize your array
double a[2][2]={{1.5 , 2.5},{3.5 , 4.5}};
//the 1st argument is the address of the first row i.e
//the first 1D array
//the 2nd argument is the no of rows of your array
//the 3rd argument is the no of columns of your array
myFunc(a[0],2,2);
//same way as myFunc
printArray(a[0],2,2);
return 0;
}
Your output will be as follows...
11.5 12.5
13.5 14.5
One important thing for passing multidimensional arrays is:
First array dimension need not be specified.
Second(any any further)dimension must be specified.
1.When only second dimension is available globally (either as a macro or as a global constant)
const int N = 3;
void print(int arr[][N], int m)
{
int i, j;
for (i = 0; i < m; i++)
for (j = 0; j < N; j++)
printf("%d ", arr[i][j]);
}
int main()
{
int arr[][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
print(arr, 3);
return 0;
}
2.Using a single pointer:
In this method,we must typecast the 2D array when passing to function.
void print(int *arr, int m, int n)
{
int i, j;
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
printf("%d ", *((arr+i*n) + j));
}
int main()
{
int arr[][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int m = 3, n = 3;
// We can also use "print(&arr[0][0], m, n);"
print((int *)arr, m, n);
return 0;
}
#include <iostream>
/**
* Prints out the elements of a 2D array row by row.
*
* #param arr The 2D array whose elements will be printed.
*/
template <typename T, size_t rows, size_t cols>
void Print2DArray(T (&arr)[rows][cols]) {
std::cout << '\n';
for (size_t row = 0; row < rows; row++) {
for (size_t col = 0; col < cols; col++) {
std::cout << arr[row][col] << ' ';
}
std::cout << '\n';
}
}
int main()
{
int i[2][5] = { {0, 1, 2, 3, 4},
{5, 6, 7, 8, 9} };
char c[3][9] = { {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'},
{'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R'},
{'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '&'} };
std::string s[4][4] = { {"Amelia", "Edward", "Israel", "Maddox"},
{"Brandi", "Fabian", "Jordan", "Norman"},
{"Carmen", "George", "Kelvin", "Oliver"},
{"Deanna", "Harvey", "Ludwig", "Philip"} };
Print2DArray(i);
Print2DArray(c);
Print2DArray(s);
std::cout <<'\n';
}
In the case you want to pass a dynamic sized 2-d array to a function, using some pointers could work for you.
void func1(int *arr, int n, int m){
...
int i_j_the_element = arr[i * m + j]; // use the idiom of i * m + j for arr[i][j]
...
}
void func2(){
...
int arr[n][m];
...
func1(&(arr[0][0]), n, m);
}
You can use template facility in C++ to do this. I did something like this :
template<typename T, size_t col>
T process(T a[][col], size_t row) {
...
}
the problem with this approach is that for every value of col which you provide, the a new function definition is instantiated using the template.
so,
int some_mat[3][3], another_mat[4,5];
process(some_mat, 3);
process(another_mat, 4);
instantiates the template twice to produce 2 function definitions (one where col = 3 and one where col = 5).
If you want to pass int a[2][3] to void func(int** pp) you need auxiliary steps as follows.
int a[2][3];
int* p[2] = {a[0],a[1]};
int** pp = p;
func(pp);
As the first [2] can be implicitly specified, it can be simplified further as.
int a[][3];
int* p[] = {a[0],a[1]};
int** pp = p;
func(pp);
You are allowed to omit the leftmost dimension and so you end up with two options:
void f1(double a[][2][3]) { ... }
void f2(double (*a)[2][3]) { ... }
double a[1][2][3];
f1(a); // ok
f2(a); // ok
This is the same with pointers:
// compilation error: cannot convert ‘double (*)[2][3]’ to ‘double***’
// double ***p1 = a;
// compilation error: cannot convert ‘double (*)[2][3]’ to ‘double (**)[3]’
// double (**p2)[3] = a;
double (*p3)[2][3] = a; // ok
// compilation error: array of pointers != pointer to array
// double *p4[2][3] = a;
double (*p5)[3] = a[0]; // ok
double *p6 = a[0][1]; // ok
The decay of an N dimensional array to a pointer to N-1 dimensional array is allowed by C++ standard, since you can lose the leftmost dimension and still being able to correctly access array elements with N-1 dimension information.
Details in here
Though, arrays and pointers are not the same: an array can decay into a pointer, but a pointer doesn't carry state about the size/configuration of the data to which it points.
A char ** is a pointer to a memory block containing character pointers, which themselves point to memory blocks of characters. A char [][] is a single memory block which contains characters. This has an impact on how the compiler translate the code and how the final performance will be.
Source
Despite appearances, the data structure implied by double** is fundamentally incompatible with that of a fixed c-array (double[][]).
The problem is that both are popular (although) misguided ways to deal with arrays in C (or C++).
See https://www.fftw.org/fftw3_doc/Dynamic-Arrays-in-C_002dThe-Wrong-Way.html
If you can't control either part of the code you need a translation layer (called adapt here), as explained here: https://c-faq.com/aryptr/dynmuldimary.html
You need to generate an auxiliary array of pointers, pointing to each row of the c-array.
#include<algorithm>
#include<cassert>
#include<vector>
void myFunction(double** myArray) {
myArray[2][3] = 5;
}
template<std::size_t N, std::size_t M>
auto adapt(double(&Carr2D)[N][M]) {
std::array<double*, N> ret;
std::transform(
std::begin(Carr2D), std::end(Carr2D),
ret.begin(),
[](auto&& row) { return &row[0];}
);
return ret;
}
int main() {
double anArray[10][10];
myFunction( adapt(anArray).data() );
assert(anArray[2][3] == 5);
}
(see working code here: https://godbolt.org/z/7M7KPzbWY)
If it looks like a recipe for disaster is because it is, as I said the two data structures are fundamentally incompatible.
If you can control both ends of the code, these days, you are better off using a modern (or semimodern) array library, like Boost.MultiArray, Boost.uBLAS, Eigen or Multi.
If the arrays are going to be small, you have "tiny" arrays libraries, for example inside Eigen or if you can't afford any dependency you might try simply with std::array<std::array<double, N>, M>.
With Multi, you can simply do this:
#include<multi/array.hpp>
#include<cassert>
namespace multi = boost::multi;
template<class Array2D>
void myFunction(Array2D&& myArray) {
myArray[2][3] = 5;
}
int main() {
multi::array<double, 2> anArray({10, 10});
myFunction(anArray);
assert(anArray[2][3] == 5);
}
(working code: https://godbolt.org/z/7M7KPzbWY)
You could take arrays of an arbitrary number of dimensions by reference and peel off one layer at a time recursively.
Here's an example of a print function for demonstrational purposes:
#include <cstddef>
#include <iostream>
#include <iterator>
#include <string>
#include <type_traits>
template <class T, std::size_t N>
void print(const T (&arr)[N], unsigned indent = 0) {
if constexpr (std::rank_v<T> == 0) {
// inner layer - print the values:
std::cout << std::string(indent, ' ') << '{';
auto it = std::begin(arr);
std::cout << *it;
for (++it; it != std::end(arr); ++it) {
std::cout << ", " << *it;
}
std::cout << '}';
} else {
// still more layers to peel off:
std::cout << std::string(indent, ' ') << "{\n";
auto it = std::begin(arr);
print(*it, indent + 1);
for (++it; it != std::end(arr); ++it) {
std::cout << ",\n";
print(*it, indent + 1);
}
std::cout << '\n' << std::string(indent, ' ') << '}';
}
}
Here's a usage example with a 3 dimensional array:
int main() {
int array[2][3][5]
{
{
{1, 2, 9, -5, 3},
{6, 7, 8, -45, -7},
{11, 12, 13, 14, 25}
},
{
{4, 5, 0, 33, 34},
{8, 9, 99, 54, 44},
{14, 15, 16, 19, 20}
}
};
print(array);
}
... which will produce this output:
{
{
{1, 2, 9, -5, 3},
{6, 7, 8, -45, -7},
{11, 12, 13, 14, 25}
},
{
{4, 5, 0, 33, 34},
{8, 9, 99, 54, 44},
{14, 15, 16, 19, 20}
}
}

c++ How to pass a 2D array to a function that expects a pointer [duplicate]

I have a function which I want to take, as a parameter, a 2D array of variable size.
So far I have this:
void myFunction(double** myArray){
myArray[x][y] = 5;
etc...
}
And I have declared an array elsewhere in my code:
double anArray[10][10];
However, calling myFunction(anArray) gives me an error.
I do not want to copy the array when I pass it in. Any changes made in myFunction should alter the state of anArray. If I understand correctly, I only want to pass in as an argument a pointer to a 2D array. The function needs to accept arrays of different sizes also. So for example, [10][10] and [5][5]. How can I do this?
There are three ways to pass a 2D array to a function:
The parameter is a 2D array
int array[10][10];
void passFunc(int a[][10])
{
// ...
}
passFunc(array);
The parameter is an array containing pointers
int *array[10];
for(int i = 0; i < 10; i++)
array[i] = new int[10];
void passFunc(int *a[10]) //Array containing pointers
{
// ...
}
passFunc(array);
The parameter is a pointer to a pointer
int **array;
array = new int *[10];
for(int i = 0; i <10; i++)
array[i] = new int[10];
void passFunc(int **a)
{
// ...
}
passFunc(array);
Fixed Size
1. Pass by reference
template <size_t rows, size_t cols>
void process_2d_array_template(int (&array)[rows][cols])
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < cols; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
In C++ passing the array by reference without losing the dimension information is probably the safest, since one needn't worry about the caller passing an incorrect dimension (compiler flags when mismatching). However, this isn't possible with dynamic (freestore) arrays; it works for automatic (usually stack-living) arrays only i.e. the dimensionality should be known at compile time.
2. Pass by pointer
void process_2d_array_pointer(int (*array)[5][10])
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < 5; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < 10; ++j)
std::cout << (*array)[i][j] << '\t';
std::cout << std::endl;
}
}
The C equivalent of the previous method is passing the array by pointer. This should not be confused with passing by the array's decayed pointer type (3), which is the common, popular method, albeit less safe than this one but more flexible. Like (1), use this method when all the dimensions of the array is fixed and known at compile-time. Note that when calling the function the array's address should be passed process_2d_array_pointer(&a) and not the address of the first element by decay process_2d_array_pointer(a).
Variable Size
These are inherited from C but are less safe, the compiler has no way of checking, guaranteeing that the caller is passing the required dimensions. The function only banks on what the caller passes in as the dimension(s). These are more flexible than the above ones since arrays of different lengths can be passed to them invariably.
It is to be remembered that there's no such thing as passing an array directly to a function in C [while in C++ they can be passed as a reference (1)]; (2) is passing a pointer to the array and not the array itself. Always passing an array as-is becomes a pointer-copy operation which is facilitated by array's nature of decaying into a pointer.
3. Pass by (value) a pointer to the decayed type
// int array[][10] is just fancy notation for the same thing
void process_2d_array(int (*array)[10], size_t rows)
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < 10; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
Although int array[][10] is allowed, I'd not recommend it over the above syntax since the above syntax makes it clear that the identifier array is a single pointer to an array of 10 integers, while this syntax looks like it's a 2D array but is the same pointer to an array of 10 integers. Here we know the number of elements in a single row (i.e. the column size, 10 here) but the number of rows is unknown and hence to be passed as an argument. In this case there's some safety since the compiler can flag when a pointer to an array with second dimension not equal to 10 is passed. The first dimension is the varying part and can be omitted. See here for the rationale on why only the first dimension is allowed to be omitted.
4. Pass by pointer to a pointer
// int *array[10] is just fancy notation for the same thing
void process_pointer_2_pointer(int **array, size_t rows, size_t cols)
{
std::cout << __func__ << std::endl;
for (size_t i = 0; i < rows; ++i)
{
std::cout << i << ": ";
for (size_t j = 0; j < cols; ++j)
std::cout << array[i][j] << '\t';
std::cout << std::endl;
}
}
Again there's an alternative syntax of int *array[10] which is the same as int **array. In this syntax the [10] is ignored as it decays into a pointer thereby becoming int **array. Perhaps it is just a cue to the caller that the passed array should have at least 10 columns, even then row count is required. In any case the compiler doesn't flag for any length/size violations (it only checks if the type passed is a pointer to pointer), hence requiring both row and column counts as parameter makes sense here.
Note: (4) is the least safest option since it hardly has any type check and the most inconvenient. One cannot legitimately pass a 2D array to this function; C-FAQ condemns the usual workaround of doing int x[5][10]; process_pointer_2_pointer((int**)&x[0][0], 5, 10); as it may potentially lead to undefined behaviour due to array flattening. The right way of passing an array in this method brings us to the inconvenient part i.e. we need an additional (surrogate) array of pointers with each of its element pointing to the respective row of the actual, to-be-passed array; this surrogate is then passed to the function (see below); all this for getting the same job done as the above methods which are more safer, cleaner and perhaps faster.
Here's a driver program to test the above functions:
#include <iostream>
// copy above functions here
int main()
{
int a[5][10] = { { } };
process_2d_array_template(a);
process_2d_array_pointer(&a); // <-- notice the unusual usage of addressof (&) operator on an array
process_2d_array(a, 5);
// works since a's first dimension decays into a pointer thereby becoming int (*)[10]
int *b[5]; // surrogate
for (size_t i = 0; i < 5; ++i)
{
b[i] = a[i];
}
// another popular way to define b: here the 2D arrays dims may be non-const, runtime var
// int **b = new int*[5];
// for (size_t i = 0; i < 5; ++i) b[i] = new int[10];
process_pointer_2_pointer(b, 5, 10);
// process_2d_array(b, 5);
// doesn't work since b's first dimension decays into a pointer thereby becoming int**
}
A modification to shengy's first suggestion, you can use templates to make the function accept a multi-dimensional array variable (instead of storing an array of pointers that have to be managed and deleted):
template <size_t size_x, size_t size_y>
void func(double (&arr)[size_x][size_y])
{
printf("%p\n", &arr);
}
int main()
{
double a1[10][10];
double a2[5][5];
printf("%p\n%p\n\n", &a1, &a2);
func(a1);
func(a2);
return 0;
}
The print statements are there to show that the arrays are getting passed by reference (by displaying the variables' addresses)
Surprised that no one mentioned this yet, but you can simply template on anything 2D supporting [][] semantics.
template <typename TwoD>
void myFunction(TwoD& myArray){
myArray[x][y] = 5;
etc...
}
// call with
double anArray[10][10];
myFunction(anArray);
It works with any 2D "array-like" datastructure, such as std::vector<std::vector<T>>, or a user defined type to maximize code reuse.
You can create a function template like this:
template<int R, int C>
void myFunction(double (&myArray)[R][C])
{
myArray[x][y] = 5;
etc...
}
Then you have both dimension sizes via R and C. A different function will be created for each array size, so if your function is large and you call it with a variety of different array sizes, this may be costly. You could use it as a wrapper over a function like this though:
void myFunction(double * arr, int R, int C)
{
arr[x * C + y] = 5;
etc...
}
It treats the array as one dimensional, and uses arithmetic to figure out the offsets of the indexes. In this case, you would define the template like this:
template<int C, int R>
void myFunction(double (&myArray)[R][C])
{
myFunction(*myArray, R, C);
}
anArray[10][10] is not a pointer to a pointer, it is a contiguous chunk of memory suitable for storing 100 values of type double, which compiler knows how to address because you specified the dimensions. You need to pass it to a function as an array. You can omit the size of the initial dimension, as follows:
void f(double p[][10]) {
}
However, this will not let you pass arrays with the last dimension other than ten.
The best solution in C++ is to use std::vector<std::vector<double> >: it is nearly as efficient, and significantly more convenient.
Here is a vector of vectors matrix example
#include <iostream>
#include <vector>
using namespace std;
typedef vector< vector<int> > Matrix;
void print(Matrix& m)
{
int M=m.size();
int N=m[0].size();
for(int i=0; i<M; i++) {
for(int j=0; j<N; j++)
cout << m[i][j] << " ";
cout << endl;
}
cout << endl;
}
int main()
{
Matrix m = { {1,2,3,4},
{5,6,7,8},
{9,1,2,3} };
print(m);
//To initialize a 3 x 4 matrix with 0:
Matrix n( 3,vector<int>(4,0));
print(n);
return 0;
}
output:
1 2 3 4
5 6 7 8
9 1 2 3
0 0 0 0
0 0 0 0
0 0 0 0
Single dimensional array decays to a pointer pointer pointing to the first element in the array. While a 2D array decays to a pointer pointing to first row. So, the function prototype should be -
void myFunction(double (*myArray) [10]);
I would prefer std::vector over raw arrays.
We can use several ways to pass a 2D array to a function:
Using single pointer we have to typecast the 2D array.
#include<bits/stdc++.h>
using namespace std;
void func(int *arr, int m, int n)
{
for (int i=0; i<m; i++)
{
for (int j=0; j<n; j++)
{
cout<<*((arr+i*n) + j)<<" ";
}
cout<<endl;
}
}
int main()
{
int m = 3, n = 3;
int arr[m][n] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
func((int *)arr, m, n);
return 0;
}
Using double pointer In this way, we also typecast the 2d array
#include<bits/stdc++.h>
using namespace std;
void func(int **arr, int row, int col)
{
for (int i=0; i<row; i++)
{
for(int j=0 ; j<col; j++)
{
cout<<arr[i][j]<<" ";
}
printf("\n");
}
}
int main()
{
int row, colum;
cin>>row>>colum;
int** arr = new int*[row];
for(int i=0; i<row; i++)
{
arr[i] = new int[colum];
}
for(int i=0; i<row; i++)
{
for(int j=0; j<colum; j++)
{
cin>>arr[i][j];
}
}
func(arr, row, colum);
return 0;
}
You can do something like this...
#include<iostream>
using namespace std;
//for changing values in 2D array
void myFunc(double *a,int rows,int cols){
for(int i=0;i<rows;i++){
for(int j=0;j<cols;j++){
*(a+ i*rows + j)+=10.0;
}
}
}
//for printing 2D array,similar to myFunc
void printArray(double *a,int rows,int cols){
cout<<"Printing your array...\n";
for(int i=0;i<rows;i++){
for(int j=0;j<cols;j++){
cout<<*(a+ i*rows + j)<<" ";
}
cout<<"\n";
}
}
int main(){
//declare and initialize your array
double a[2][2]={{1.5 , 2.5},{3.5 , 4.5}};
//the 1st argument is the address of the first row i.e
//the first 1D array
//the 2nd argument is the no of rows of your array
//the 3rd argument is the no of columns of your array
myFunc(a[0],2,2);
//same way as myFunc
printArray(a[0],2,2);
return 0;
}
Your output will be as follows...
11.5 12.5
13.5 14.5
One important thing for passing multidimensional arrays is:
First array dimension need not be specified.
Second(any any further)dimension must be specified.
1.When only second dimension is available globally (either as a macro or as a global constant)
const int N = 3;
void print(int arr[][N], int m)
{
int i, j;
for (i = 0; i < m; i++)
for (j = 0; j < N; j++)
printf("%d ", arr[i][j]);
}
int main()
{
int arr[][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
print(arr, 3);
return 0;
}
2.Using a single pointer:
In this method,we must typecast the 2D array when passing to function.
void print(int *arr, int m, int n)
{
int i, j;
for (i = 0; i < m; i++)
for (j = 0; j < n; j++)
printf("%d ", *((arr+i*n) + j));
}
int main()
{
int arr[][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
int m = 3, n = 3;
// We can also use "print(&arr[0][0], m, n);"
print((int *)arr, m, n);
return 0;
}
#include <iostream>
/**
* Prints out the elements of a 2D array row by row.
*
* #param arr The 2D array whose elements will be printed.
*/
template <typename T, size_t rows, size_t cols>
void Print2DArray(T (&arr)[rows][cols]) {
std::cout << '\n';
for (size_t row = 0; row < rows; row++) {
for (size_t col = 0; col < cols; col++) {
std::cout << arr[row][col] << ' ';
}
std::cout << '\n';
}
}
int main()
{
int i[2][5] = { {0, 1, 2, 3, 4},
{5, 6, 7, 8, 9} };
char c[3][9] = { {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I'},
{'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R'},
{'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '&'} };
std::string s[4][4] = { {"Amelia", "Edward", "Israel", "Maddox"},
{"Brandi", "Fabian", "Jordan", "Norman"},
{"Carmen", "George", "Kelvin", "Oliver"},
{"Deanna", "Harvey", "Ludwig", "Philip"} };
Print2DArray(i);
Print2DArray(c);
Print2DArray(s);
std::cout <<'\n';
}
In the case you want to pass a dynamic sized 2-d array to a function, using some pointers could work for you.
void func1(int *arr, int n, int m){
...
int i_j_the_element = arr[i * m + j]; // use the idiom of i * m + j for arr[i][j]
...
}
void func2(){
...
int arr[n][m];
...
func1(&(arr[0][0]), n, m);
}
You can use template facility in C++ to do this. I did something like this :
template<typename T, size_t col>
T process(T a[][col], size_t row) {
...
}
the problem with this approach is that for every value of col which you provide, the a new function definition is instantiated using the template.
so,
int some_mat[3][3], another_mat[4,5];
process(some_mat, 3);
process(another_mat, 4);
instantiates the template twice to produce 2 function definitions (one where col = 3 and one where col = 5).
If you want to pass int a[2][3] to void func(int** pp) you need auxiliary steps as follows.
int a[2][3];
int* p[2] = {a[0],a[1]};
int** pp = p;
func(pp);
As the first [2] can be implicitly specified, it can be simplified further as.
int a[][3];
int* p[] = {a[0],a[1]};
int** pp = p;
func(pp);
You are allowed to omit the leftmost dimension and so you end up with two options:
void f1(double a[][2][3]) { ... }
void f2(double (*a)[2][3]) { ... }
double a[1][2][3];
f1(a); // ok
f2(a); // ok
This is the same with pointers:
// compilation error: cannot convert ‘double (*)[2][3]’ to ‘double***’
// double ***p1 = a;
// compilation error: cannot convert ‘double (*)[2][3]’ to ‘double (**)[3]’
// double (**p2)[3] = a;
double (*p3)[2][3] = a; // ok
// compilation error: array of pointers != pointer to array
// double *p4[2][3] = a;
double (*p5)[3] = a[0]; // ok
double *p6 = a[0][1]; // ok
The decay of an N dimensional array to a pointer to N-1 dimensional array is allowed by C++ standard, since you can lose the leftmost dimension and still being able to correctly access array elements with N-1 dimension information.
Details in here
Though, arrays and pointers are not the same: an array can decay into a pointer, but a pointer doesn't carry state about the size/configuration of the data to which it points.
A char ** is a pointer to a memory block containing character pointers, which themselves point to memory blocks of characters. A char [][] is a single memory block which contains characters. This has an impact on how the compiler translate the code and how the final performance will be.
Source
Despite appearances, the data structure implied by double** is fundamentally incompatible with that of a fixed c-array (double[][]).
The problem is that both are popular (although) misguided ways to deal with arrays in C (or C++).
See https://www.fftw.org/fftw3_doc/Dynamic-Arrays-in-C_002dThe-Wrong-Way.html
If you can't control either part of the code you need a translation layer (called adapt here), as explained here: https://c-faq.com/aryptr/dynmuldimary.html
You need to generate an auxiliary array of pointers, pointing to each row of the c-array.
#include<algorithm>
#include<cassert>
#include<vector>
void myFunction(double** myArray) {
myArray[2][3] = 5;
}
template<std::size_t N, std::size_t M>
auto adapt(double(&Carr2D)[N][M]) {
std::array<double*, N> ret;
std::transform(
std::begin(Carr2D), std::end(Carr2D),
ret.begin(),
[](auto&& row) { return &row[0];}
);
return ret;
}
int main() {
double anArray[10][10];
myFunction( adapt(anArray).data() );
assert(anArray[2][3] == 5);
}
(see working code here: https://godbolt.org/z/7M7KPzbWY)
If it looks like a recipe for disaster is because it is, as I said the two data structures are fundamentally incompatible.
If you can control both ends of the code, these days, you are better off using a modern (or semimodern) array library, like Boost.MultiArray, Boost.uBLAS, Eigen or Multi.
If the arrays are going to be small, you have "tiny" arrays libraries, for example inside Eigen or if you can't afford any dependency you might try simply with std::array<std::array<double, N>, M>.
With Multi, you can simply do this:
#include<multi/array.hpp>
#include<cassert>
namespace multi = boost::multi;
template<class Array2D>
void myFunction(Array2D&& myArray) {
myArray[2][3] = 5;
}
int main() {
multi::array<double, 2> anArray({10, 10});
myFunction(anArray);
assert(anArray[2][3] == 5);
}
(working code: https://godbolt.org/z/7M7KPzbWY)
You could take arrays of an arbitrary number of dimensions by reference and peel off one layer at a time recursively.
Here's an example of a print function for demonstrational purposes:
#include <cstddef>
#include <iostream>
#include <iterator>
#include <string>
#include <type_traits>
template <class T, std::size_t N>
void print(const T (&arr)[N], unsigned indent = 0) {
if constexpr (std::rank_v<T> == 0) {
// inner layer - print the values:
std::cout << std::string(indent, ' ') << '{';
auto it = std::begin(arr);
std::cout << *it;
for (++it; it != std::end(arr); ++it) {
std::cout << ", " << *it;
}
std::cout << '}';
} else {
// still more layers to peel off:
std::cout << std::string(indent, ' ') << "{\n";
auto it = std::begin(arr);
print(*it, indent + 1);
for (++it; it != std::end(arr); ++it) {
std::cout << ",\n";
print(*it, indent + 1);
}
std::cout << '\n' << std::string(indent, ' ') << '}';
}
}
Here's a usage example with a 3 dimensional array:
int main() {
int array[2][3][5]
{
{
{1, 2, 9, -5, 3},
{6, 7, 8, -45, -7},
{11, 12, 13, 14, 25}
},
{
{4, 5, 0, 33, 34},
{8, 9, 99, 54, 44},
{14, 15, 16, 19, 20}
}
};
print(array);
}
... which will produce this output:
{
{
{1, 2, 9, -5, 3},
{6, 7, 8, -45, -7},
{11, 12, 13, 14, 25}
},
{
{4, 5, 0, 33, 34},
{8, 9, 99, 54, 44},
{14, 15, 16, 19, 20}
}
}

How do I stop my array from printing address?

I have a header, cpp and main class.
//Arr.h
class Arr
{
public:
void setArr();
void printArr();
private:
int x[5];
};
//Arr.cpp
#include "Arr.h"
#include <iostream>
using namespace std;
void Arr::setArr()
{
int x[5] = { 2, 3, 5, 7, 11 };
}
void Arr::printArr()
{
for (int i = 0; i < 5; i++)
{
cout << x[i] << "\n";
}
}
//main.cpp
int main()
{
Arr a;
a.setArr();
a.printArr();
}
However, when I run the code, a.printArr() prints out array address and not the values contained in the array. Is there a way to fix this?
Your code will print not address but some indeterminate value generated via default-initializing. Initialize the member array instead of the local array to throw away.
void Arr::setArr()
{
x[0] = 2;
x[1] = 3;
x[2] = 5;
x[3] = 7;
x[4] = 11;
}
Your problem is here:
void Arr::setArr()
{
int x[5] = { 2, 3, 5, 7, 11 };
}
the declaration int x[5] defines a new array of 5 elements which will be destroyed on exiting that function; and the name x hides the data member Arr::x.
One way you can do it is this:
void Arr::setArr()
{
/*static*/ auto arr = { 2, 3, 5, 7, 11 };
std::copy(arr.begin(), arr.end(), x);
}
Full code:
#include <iostream>
#include <algorithm>
class Arr
{
public:
void setArr();
void printArr();
private:
int x[5];
};
using namespace std;
void Arr::setArr()
{
/*static*/ auto arr = { 2, 3, 5, 7, 11 };
std::copy(arr.begin(), arr.end(), x);
}
void Arr::printArr()
{
for (int i = 0; i < 5; i++)
{
cout << x[i] << "\n";
}
}
//main.cpp
int main()
{
Arr a;
a.setArr();
a.printArr();
}
Your code is not printing array address. Maybe you saw some unexpected numbers and assumed it was an address, but in fact it is undefined behaviour caused by outputting uninitialized variables.
Your code int x[5] = .... failed because this declares a new local variable x, it doesn't modify the class member x.
Perhaps you tried:
x = { 2, 3, 5, 7, 11 };
and got compilation errors. This is because C-style arrays may not be assigned. This annoying rule is a hangover from C++'s past. To avoid this problem and many others, you can try to avoid using C-style arrays. In this case use a C++ array:
private:
std::array<int, 5> x;
Then the suggested assignment will work.
As WhiZTiM has already pointed out, your problem is at the setArr function in your class. The reason for this is because in C++ you cannot assign values to the array in the "normal fashion" i.e. using x = {blah, blah, blah}; (unless you use std::array<int, ARRAYSIZE>). In setArr you are creating another array named x and then this array is deleted once it is out of scope, and in printArr you are printing an array with no values in it yet.
Each value in the array must be set individually in C styled arrays (as shown in MikeCAT's answer).
One solution to this is to create a temporary array with the values you want and assigning the values to your class array through a for loop, i.e.:
void Arr::setArr() {
const int ARRAYSIZE = 5;
int tmp[ARRAYSIZE] = {2, 3, 5, 7, 11};
for (int i = 0; i < ARRAYSIZE; ++i) {
x[i] = tmp[i];
}
}
As M.M also pointed out, can simply change int x[5] in your private variables in Arr to std::array<int, 5> x and then in setArr simply have the statement: x = {2, 3, 5, 7, 11}; which is the better option in my opinion and easier to read, but it's also good to know how to use C arrays 😊