Using arithmetic with arrays - c++

Two quantities u and v are said to be at right angles if
nuv = u1v1 + u2v2 + u3v3 + u4v4 +………… + unvn = 0
Write a function that computes whether u and v are at right angles. You may use arrays if you wish. The function can assume that the vectors have the same dimension (n, say), but this number should be passed as a parameter to the function.
I have a few errors in my program. I'd appreciate some help, as I'm a beginner. The errors are telling me:
In function 'void function(int*,int*)'
cpp 26: error expected ';' before '}' token
cpp 29: error ;value required as left operand of assignment
#include <iostream>
using namespace std;
const int n = 5;
void function(int array[n],int array2[n]);
int main(){
int array[n] = {5, 3 , -4, 2, 8};
int array2[n] ={-7, -9, 5, 2, 9};
function(array, array2);
return 0;
}
void function(int array[n], int array2[n]){
int multiple;
for(int i=0; i <=5, i++)
{
(array[i]*array2[i]) + (array[i+1]*array2[i+1]) = multiple;
}
cout << multiple << endl;
}

Your for loop is malformed. You need to use < instead of <=, use n instead of 5, and use ; instead of ,.
Your assignment of multiple is backwards. The value on the right-side of the = operator is assigned to the variable on the left-side of =. You are trying to assign the value of multiple (which is uninitialized) to a dynamically computed value that has no explicit variable of its own. You should be assigning the computed value to multiple instead.
Also, you didn't follow the "this number [array dimensions] should be passed as a parameter to the function" requirement of your instructions.
Try this:
#include <iostream>
using namespace std;
const int n = 5;
void function(const int *array1, const int *array2, const int size);
int main()
{
int array1[n] = { 5, 3, -4, 2, 8};
int array2[n] = {-7, -9, 5, 2, 9};
function(array1, array2, n);
return 0;
}
void function(const int *array1, const int *array2, const int size)
{
int multiple = 0;
for(int i = 0; i < size; i++)
{
multiple += (array1[i] * array2[i]);
}
cout << multiple << endl;
}

Syntax error is where you are using the for loop in this fashion:
for(int i=0;i<=5,i++)
Use this instead:
for(int i=0; i <= 5; i++)

The function can assume that the vectors have the same dimension (n,
say), but this number should be passed as a parameter to the
function.
This function declaration
void function(int array[n],int array2[n]);
does not include a parameter that specifies the dimension of the arrays.
The above declaration is equivalent to
void function(int *array,int *array2);
because arrays passed by value are implicitly converted to pointers to their first elements.
There are a typo and a bug in this for statement
for(int i=0; i <=5, i++)
^^^^^^
There shall be
for ( int i=0; i < n; i++)
The variable multiple
int multiple;
is not initialized and this assignment
(array[i]*array2[i]) + (array[i+1]*array2[i+1]) = multiple;
does not have sense and has nothing common with the condition
nuv = u1v1 + u2v2 + u3v3 + u4v4 +………… + unvn = 0
It seems what you mean is the following
#include <iostream>
bool function( const int array[], const int array2[], size_t n )
{
long long int product = 0;
for ( size_t i = 0; i < n; i++)
{
product += array[i] * array2[i];
}
return product == 0;
}
int main()
{
const size_t N = 5;
int array[N] = { 5, 3 , -4, 2, 8 };
int array2[N] = { -7, -9, 5, 2, 9 };
std::cout << "array and array2 are "
<< (function(array, array2, N) ? "" : "not ")
<< "at right angles"
<< std::endl;
return 0;
}
These arrays
int array[N] = { 5, 3 , -4, 2, 8 };
int array2[N] = { -7, -9, 5, 2, 9 };
are not a right angles,
But these arrays
int array[N] = { 5, 3 , -4, 1, 9 };
int array2[N] = { -7, -9, 5, 1, 9 };
are at right angles. Try them.

Alternative: try the C++ way. Use std::array that knows it's length. Use algorithms provided by the standard library like std::inner_product.
#include <iostream>
#include <algorithm>
#include <array>
#include <numeric>
int main()
{
using arr_t = std::array<int,5>;
arr_t arr1 = {5, 3 , -4, 2, 8};
arr_t arr2 = {-7, -9, 5, 2, 9};
int mult = std::inner_product( begin(arr1), end(arr1), begin(arr2), 0,
std::plus<>(), std::multiplies<>() );
std::cerr << mult << "\n";
}

Related

How do I linear search two arrays in c++?

I have two arrays of data type double - called array1[10] and array2[8]. I am required to search for array2 inside array1 at each element using a linear search function. The function declaration is
string linSearch (double array1[10], double array2[8]);
If array2 is found inside array1 then I need to print out the index of where it is found in array1. If its not found I need the output to be "NA". This output must be a delimited-comma- string.
eg.
//given the two arrays:
array1={1.1,1.2,6,7,3.5,2,7,8.8,9,23.4}
array2={6,45,2,7,1.1,5,4,8.8}
//after the linear search completes, the output must be the index in which //array2 is found in array1. if its not found, then it must be NA:
2,NA,5,6,0,NA,NA,7
So far I have the code that follows. Its my first time working with arrays and I am still having difficulties grasping the concept- like once I define the function how do I even call it in the main program?! anyway..the function definition I have (excluding the main program) is:
string linSearch (double array1[10], double array2[8])
{
int index1 = 0;
int index2 =0;
int position =-1;
bool found = false;
while (index1<10 && !found && index2<8)
{
if array1[index1] == array2[index2])
{
found = true;
position = index1;
}
index1++;
index2++;
}
return position;
}
I am EXTREMELY confused about searching for one array in the other and how to output the delimited list as well as how to connect it to my main program. Any help would be GREATLY appreciated. Thanks!
#include <iostream>
using namespace std;
string linSearch(double array1[10], double array2[8])
{
string result = "";
bool found;
for (int j = 0; j < 8; j++)
{
if(j > 0)
result.append(", ");
found = false;
for (int i = 0; i < 10; i++){
if (array1[i] == array2[j]) {
result.append(to_string(i));
found = true;
}
}
if(!found)
result.append("NA");
}
return result;
}
int main(){
double a1[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
double a2[8] = {11, 25, 3, 41, 5, 6, 7, 8};
cout << linSearch(a1, a2) << endl;
return 0;
}
You are not searching one array inside the other. You are searching for elements from one array in a second array. If you are using a linear search and if you do not want to sort the array, you need 2 nested loops to do that. One for each element in the second array, and one to find that element in the first array.
Keep things simple and start with finding the position of a single element in one array. Because you are comparing doubles, you should not compare them with ==. Next you just need a function that calls the first for each element in the second array:
#include <string>
#include <vector>
#include <algorithm>
#include <iostream>
void index_to_string(const std::vector<double>& v,double e,std::ostream& out){
auto it = std::find_if(v.begin(),
v.end(),
[e](const double& x) {
return std::abs(e-x) < 1e-8;
}
);
if (it == v.end()) {
out << "NA";
} else {
out << (it - v.begin());
}
}
void all_indices_to_string(const std::vector<double>& v1,const std::vector<double>& v2,std::ostream& out){
if (v1.size() == 0 || v2.size()==0) return;
index_to_string(v1,v2[0],out);
for (size_t i=1;i<v2.size();++i){
out << ",";
index_to_string(v1,v2[i],out);
}
}
int main() {
double array1[] ={1.1,1.2,6,7,3.5,2,7,8.8,9,23.4};
double array2[] ={6,45,2,7,1.1,5,4,8.8};
all_indices_to_string(
std::vector<double>(std::begin(array1),std::end(array1)),
std::vector<double>(std::begin(array2),std::end(array2)),
std::cout
);
}
Output:
2,NA,5,3,0,NA,NA,7
In your example of arrays and the expected output
//given the two arrays:
array1={1.1,1.2,6,7,3.5,2,7,8.8,9,23.4}
array2={6,45,2,7,1.1,5,4,8.8}
and
2,NA,5,6,0,NA,NA,7
^
there is a typo. The output should be
2,NA,5,3,0,NA,NA,7
^
because the number 7 is found in the third position of the array array1.
Here you are.
#include <iostream>
#include <string>
#include <sstream>
std::string linearSearch( const double a1[], size_t n1, const double a2[], size_t n2 )
{
std::ostringstream os;
for ( size_t i = 0; i < n2; i++ )
{
if ( i != 0 ) os << ',';
size_t j = 0;
while ( j < n1 && a2[i] != a1[j] ) ++j;
if ( j == n1 ) os << "NA";
else os << j;
}
return os.str();
}
int main()
{
double a1[] = { 1.1, 1.2, 6, 7, 3.5, 2, 7, 8.8, 9, 23.4 };
const size_t N1 = sizeof( a1 ) / sizeof( *a1 );
double a2[] = { 6, 45, 2, 7, 1.1, 5, 4, 8.8 };
const size_t N2 = sizeof( a2 ) / sizeof( *a2 );
std::cout << linearSearch( a1, N1, a2, N2 ) << '\n';
return 0;
}
The program output is
2,NA,5,3,0,NA,NA,7

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}
}
}

array copy reversal in c++

Is there a way to copy an array to another way in reverse order by using a while loop in c++??
I'm pretty sure I know how to do one with a for loop, but I'm curious if anyone knows of a way by using a while loop
Why not something like this?
#include <algorithm>
int src[] = {1, 2, 3, 4, 5};
int dst[5];
std::reverse_copy(src, src+5, dst);
int anArray = {1, 2, 3, 4, 5};
int reverseArray[5];
int count = 4;
int place = 0;
while(place < 5) {
reverseArray[place] = anArray[count];
count--;
place++;
}
As you said that you have done using for loop, you can follow following steps to convert it to while loop.
for(int i = sizeof(arr) - 1; i >= 0; i--)
{
// your logic
}
now convert it to,
int i = sizeof(arr);
for(; i >= 0; )
{
// your logic
i--;
}
simply replace for with while and remove ; within the braces.
int i = sizeof(arr);
while(i >= 0)
{
// your logic
i--;
}
You can use std::reverse for reversing the same array and std::reverse_copy for reversing to another output array as:
int main() {
int a[]= {1,2,3,4,5,6,7,8,9,10};
const int size = sizeof(a)/sizeof(int);
int b[size];
//copying reverse to another array
reverse_copy(a, a + size, b);
cout << "b = {";
copy(b, b + size, ostream_iterator<int>(cout, ", "));
cout << "}" << endl;
//reverse the same array
reverse(a, a + size);
cout << "a = {";
copy(a, a + size, ostream_iterator<int>(cout, ", "));
cout << "}" << endl;
return 0;
}
Output:
b = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1, }
a = {10, 9, 8, 7, 6, 5, 4, 3, 2, 1, }
Demo : http://www.ideone.com/Fe5uj
There's been a few questions similar to this recently. I wonder if it is homework or an interview question somewhere. Here's one answer:
#define ELEMENT_COUNT(a) (sizeof((a))/sizeof((a)[0]))
int anArray[] = { 1, 2, 3, 4, 5 };
int reverseArray[ELEMENT_COUNT(anArray)];
int n = ELEMENT_COUNT(anArray);
int i = 0;
while(n--)
reverseArray[i++] = anArray[n];
I think it might be probing to see if you understand when expression like i++ and n-- are evaluated.

How do I find a particular value in an array and return its index?

Pseudo Code:
int arr[ 5 ] = { 4, 1, 3, 2, 6 }, x;
x = find(3).arr ;
x would then return 2.
The syntax you have there for your function doesn't make sense (why would the return value have a member called arr?).
To find the index, use std::distance and std::find from the <algorithm> header.
int x = std::distance(arr, std::find(arr, arr + 5, 3));
Or you can make it into a more generic function:
template <typename Iter>
size_t index_of(Iter first, Iter last, typename const std::iterator_traits<Iter>::value_type& x)
{
size_t i = 0;
while (first != last && *first != x)
++first, ++i;
return i;
}
Here, I'm returning the length of the sequence if the value is not found (which is consistent with the way the STL algorithms return the last iterator). Depending on your taste, you may wish to use some other form of failure reporting.
In your case, you would use it like so:
size_t x = index_of(arr, arr + 5, 3);
Here is a very simple way to do it by hand. You could also use the <algorithm>, as Peter suggests.
#include <iostream>
int find(int arr[], int len, int seek)
{
for (int i = 0; i < len; ++i)
{
if (arr[i] == seek) return i;
}
return -1;
}
int main()
{
int arr[ 5 ] = { 4, 1, 3, 2, 6 };
int x = find(arr,5,3);
std::cout << x << std::endl;
}
The fancy answer:
Use std::vector and search with std::find
The simple answer
Use a for loop
If the array is unsorted, you will need to use linear search.
#include <vector>
#include <algorithm>
int main()
{
int arr[5] = {4, 1, 3, 2, 6};
int x = -1;
std::vector<int> testVector(arr, arr + sizeof(arr) / sizeof(int) );
std::vector<int>::iterator it = std::find(testVector.begin(), testVector.end(), 3);
if (it != testVector.end())
{
x = it - testVector.begin();
}
return 0;
}
Or you can just build a vector in a normal way, without creating it from an array of ints and then use the same solution as shown in my example.
int arr[5] = {4, 1, 3, 2, 6};
vector<int> vec;
int i =0;
int no_to_be_found;
cin >> no_to_be_found;
while(i != 4)
{
vec.push_back(arr[i]);
i++;
}
cout << find(vec.begin(),vec.end(),no_to_be_found) - vec.begin();
We here use simply linear search. At first initialize the index equal to -1 . Then search the array , if found the assign the index value in index variable and break. Otherwise, index = -1.
int find(int arr[], int n, int key)
{
int index = -1;
for(int i=0; i<n; i++)
{
if(arr[i]==key)
{
index=i;
break;
}
}
return index;
}
int main()
{
int arr[ 5 ] = { 4, 1, 3, 2, 6 };
int n = sizeof(arr)/sizeof(arr[0]);
int x = find(arr ,n, 3);
cout<<x<<endl;
return 0;
}
You could use the STL algorithm library's find function provided
#include <iostream>
#include <algorithm>
using std::iostream;
using std::find;
int main() {
int length = 10;
int arr[length] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int* found_pos = find(arr, arr + length, 5);
if(found_pos != (arr + length)) {
// found
cout << "Found: " << *found_pos << endl;
}
else {
// not found
cout << "Not Found." << endl;
}
return 0;
}
There is a find(...) function to find an element in an array which returns an iterator to that element. If the element is not found, the iterator point to the end of array.
In case the element is found, we can simply calculate the distance of the iterator from the beginning of the array to get the index of that element.
#include <iterator>
using namespace std;
int arr[ 5 ] = { 4, 1, 3, 2, 6 }
auto it = arr.find(begin(arr), end(arr), 3)
if(it != end(arr))
cerr << "Found at index: " << (it-begin(arr)) << endl;
else
cerr << "Not found\n";