Is there a way of printing arrays in C++?
I'm trying to make a function that reverses a user-input array and then prints it out. I tried Googling this problem and it seemed like C++ can't print arrays. That can't be true can it?
Is there a way of printing arrays in C++?
I'm trying to make a function that reverses a user-input array and then prints it out. I tried Googling this problem and it seemed like C++ can't print arrays. That can't be true can it?
Just iterate over the elements. Like this:
for (int i = numElements - 1; i >= 0; i--)
cout << array[i];
Note: As Maxim Egorushkin pointed out, this could overflow. See his comment below for a better solution.
Use the STL
#include <iostream>
#include <vector>
#include <algorithm>
#include <iterator>
#include <ranges>
int main()
{
std::vector<int> userInput;
// Read until end of input.
// Hit control D
std::copy(std::istream_iterator<int>(std::cin),
std::istream_iterator<int>(),
std::back_inserter(userInput)
);
// ITs 2021 now move this up as probably the best way to do it.
// Range based for is now "probably" the best alternative C++20
// As we have all the range based extension being added to the language
for(auto const& value: userInput)
{
std::cout << value << ",";
}
std::cout << "\n";
// Print the array in reverse using the range based stuff
for(auto const& value: userInput | std::views::reverse)
{
std::cout << value << ",";
}
std::cout << "\n";
// Print in Normal order
std::copy(userInput.begin(),
userInput.end(),
std::ostream_iterator<int>(std::cout,",")
);
std::cout << "\n";
// Print in reverse order:
std::copy(userInput.rbegin(),
userInput.rend(),
std::ostream_iterator<int>(std::cout,",")
);
std::cout << "\n";
}
May I suggest using the fish bone operator?
for (auto x = std::end(a); x != std::begin(a); )
{
std::cout <<*--x<< ' ';
}
(Can you spot it?)
Besides the for-loop based solutions, you can also use an ostream_iterator<>. Here's an example that leverages the sample code in the (now retired) SGI STL reference:
#include <iostream>
#include <iterator>
#include <algorithm>
int main()
{
short foo[] = { 1, 3, 5, 7 };
using namespace std;
copy(foo,
foo + sizeof(foo) / sizeof(foo[0]),
ostream_iterator<short>(cout, "\n"));
}
This generates the following:
./a.out
1
3
5
7
However, this may be overkill for your needs. A straight for-loop is probably all that you need, although litb's template sugar is quite nice, too.
Edit: Forgot the "printing in reverse" requirement. Here's one way to do it:
#include <iostream>
#include <iterator>
#include <algorithm>
int main()
{
short foo[] = { 1, 3, 5, 7 };
using namespace std;
reverse_iterator<short *> begin(foo + sizeof(foo) / sizeof(foo[0]));
reverse_iterator<short *> end(foo);
copy(begin,
end,
ostream_iterator<short>(cout, "\n"));
}
and the output:
$ ./a.out
7
5
3
1
Edit: C++14 update that simplifies the above code snippets using array iterator functions like std::begin() and std::rbegin():
#include <iostream>
#include <iterator>
#include <algorithm>
int main()
{
short foo[] = { 1, 3, 5, 7 };
// Generate array iterators using C++14 std::{r}begin()
// and std::{r}end().
// Forward
std::copy(std::begin(foo),
std::end(foo),
std::ostream_iterator<short>(std::cout, "\n"));
// Reverse
std::copy(std::rbegin(foo),
std::rend(foo),
std::ostream_iterator<short>(std::cout, "\n"));
}
There are declared arrays and arrays that are not declared, but otherwise created, particularly using new
:
int *p = new int[3];
That array with 3 elements is created dynamically (and that 3
could have been calculated at runtime, too), and a pointer to it which has the size erased from its type is assigned to p
. You cannot get the size anymore to print that array. A function that only receives the pointer to it can thus not print that array.
Printing declared arrays is easy. You can use sizeof
to get their size and pass that size along to the function including a pointer to that array's elements. But you can also create a template that accepts the array, and deduces its size from its declared type:
template<typename Type, int Size>
void print(Type const(& array)[Size]) {
for(int i=0; i<Size; i++)
std::cout << array[i] << std::endl;
}
The problem with this is that it won't accept pointers (obviously). The easiest solution, I think, is to use std::vector
. It is a dynamic, resizable "array" (with the semantics you would expect from a real one), which has a size
member function:
void print(std::vector<int> const &v) {
std::vector<int>::size_type i;
for(i = 0; i<v.size(); i++)
std::cout << v[i] << std::endl;
}
You can, of course, also make this a template to accept vectors of other types.
Most of the libraries commonly used in C++ can't print arrays, per se. You'll have to loop through it manually and print out each value.
Printing arrays and dumping many different kinds of objects is a feature of higher level languages.
It certainly is! You'll have to loop through the array and print out each item individually.
This might help //Printing The Array
for (int i = 0; i < n; i++)
{cout << numbers[i];}
n is the size of the array
std::string ss[] = { "qwerty", "asdfg", "zxcvb" };
for ( auto el : ss ) std::cout << el << '\n';
Works basically like foreach.
My simple answer is:
#include <iostream>
using namespace std;
int main()
{
int data[]{ 1, 2, 7 };
for (int i = sizeof(data) / sizeof(data[0])-1; i >= 0; i--) {
cout << data[i];
}
return 0;
}
You can use reverse iterators to print an array in reverse:
#include <iostream>
int main() {
int x[] = {1,2,3,4,5};
for (auto it = std::rbegin(x); it != std::rend(x); ++it)
std::cout << *it;
}
54321
If you already reversed the array, you can replace std::rbegin
and std::rend
with std::begin
/std::end
, respectively, to iterate the array in forward direction.
It's quite straightforward to copy the array's elements to a suitable output iterator. For example (using C++20 for the Ranges version):
#include <algorithm>
#include <array>
#include <iostream>
#include <iterator>
template<typename T, std::size_t N>
std::ostream& print_array(std::ostream& os, std::array<T,N> const& arr)
{
std::ranges::copy(arr, std::ostream_iterator<T>(os, ", "));
return os;
}
Quick demo:
int main()
{
std::array example{ "zero", "one", "two", "three", };
print_array(std::cout, example) << '\n';
}
Of course it's more useful if we can output any kind of collection, not only arrays:
#include <algorithm>
#include <iterator>
#include <iosfwd>
#include <ranges>
template<std::ranges::input_range R>
std::ostream& print_array(std::ostream& os, R const& arr)
{
using T = std::ranges::range_value_t<R>;
std::ranges::copy(arr, std::ostream_iterator<T>(os, ", "));
return os;
}
The question mentions reversing the array for printing. That's easily achieved by using a view adapter:
print_array(std::cout, example | std::views::reverse) << '\n';
// Just do this, use a vector with this code and you're good lol -Daniel
#include <Windows.h>
#include <iostream>
#include <vector>
using namespace std;
int main()
{
std::vector<const char*> arry = { "Item 0","Item 1","Item 2","Item 3" ,"Item 4","Yay we at the end of the array"};
if (arry.size() != arry.size() || arry.empty()) {
printf("what happened to the array lol\n ");
system("PAUSE");
}
for (int i = 0; i < arry.size(); i++)
{
if (arry.max_size() == true) {
cout << "Max size of array reached!";
}
cout << "Array Value " << i << " = " << arry.at(i) << endl;
}
}
If you want to make a function that prints every single element in an array;
#include <iostream>
using namespace std;
int myArray[] = {1,2,3,4, 77, 88};
void coutArr(int *arr, int size){
for(int i=0; i<size/4; i++){
cout << arr[i] << endl;
}
}
int main(){
coutArr(myArray, sizeof(myArray));
}
The function above prints every single element in an array only, not commas etc.
You may be wondering "Why sizeoff(arr) divided by 4?". It's because cpp prints 4 if there's only a single element in an array.