c++ why dont print all elements when exponent

Question

I want to create an array and raise 2 to every element in that array and store it as new array arr2. Here is my code

#include <iostream>

using namespace std;

int main(){
  int arr1[7] = {1, 2, 3, 4, 5, 6, 7};

  auto arr2 = 2 ** arr1;

  cout << arr2 << endl;
}

But, it prints only the first element, it does not print the whole array. Why? So, basicaly, what I did here is I created arr1 with elements {1,2,3,4,5,6,7} and then I want arr2 to be

[2, 4, 8, 16, 32, 64, 128]

but for some reason it prints only the first element of array, it prints 2, but I want it to print all elements. Notice that 2 ** arr1 is the line where I am raising 2 to power (using exponentiation operator, i think it is how you call it if I'm not wrong) and then it should store array at array2.

What is wrong and why does it print only the first element instead all the elements?

There is no exponent operator in C++. Why are you guessing at syntax? — StoryTeller - Unslander Monica, Dec 27 '17 at 10:52
You need one or more of [these](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). — StoryTeller - Unslander Monica, Dec 27 '17 at 10:53
@StoryTeller. Thank you sir for explaining. I will learn it definitely. — , Dec 27 '17 at 10:54

Bathsheba · Answer 1 · 2017-12-27T10:55:59.450

7

** is not an exponentation operator. C++ is not Fortran.

You have multiplied 2 by the first element of arr: your statement is equivalent to int arr2 = 2 * arr1[0];. What you have entered is perfectly legal C++ (consisting of multiplication and pointer dereference), and the use of auto is adding to the obfuscation.

edited Dec 27 '17 at 10:55

answered Dec 27 '17 at 10:52

Bathsheba

231,907
34
361
483

thank you, it seems that you are right, I always wondered why `**` doesnt work for two constants – Dec 27 '17 at 10:54
3

This should join the exclusive club owned by the `-->` "operator". – StoryTeller - Unslander Monica Dec 27 '17 at 10:55
2

@StoryTeller : Also the "convert to bool" operator: `!!` – Martin Bonner supports Monica Dec 27 '17 at 11:12

score 1 · Accepted Answer · answered Dec 27 '17 at 11:19

This statement

auto arr2 = 2 ** arr1;

is equivalent to

auto arr2 = 2 * *arr1;

Array designators in expressions are converted (with rare exceptions) to pointers to their first elements.

So the above statement can be rewritten like

auto arr2 = 2 * *( &arr1[0] );

As the first element of the array arr1 is equal to 1 then you have

auto arr2 = 2 * 1;

The expression 2 * 1 has the type int.

So as result the statement can be just rewritten like

int arr2 = 2;

There is no such operator as ** in C++. Instead you have to use standard function pow.

You can do the task either by writing an appropriate loop manually or using for example the standard algorithm std::transform.

Here is a demonstrative program

#include <iostream>
#include <cmath>
#include <algorithm>
#include <iterator>

int main()
{
    int arr1[] = { 1, 2, 3, 4, 5, 6, 7 };
    int arr2[sizeof( arr1 ) / sizeof( *arr1 )];

    std::transform(std::begin(arr1), std::end(arr1),
        std::begin(arr2),
        [](int x) { return pow( 2, x ); });

    for (int x : arr1) std::cout << x << ' ';
    std::cout << std::endl;
    for (int x : arr2) std::cout << x << ' ';
    std::cout << std::endl;

    return 0;
}

Its output is

1 2 3 4 5 6 7 
2 4 8 16 32 64 128

score 0 · Answer 3 · answered Dec 27 '17 at 11:53

0

The simplest way to do what you want can be something like this.

#include<iostream>
#include<math.h>
using namespace std;
int main(){
      int arr1[7] = {1,2,3,4,5,6,7}, arr2[7];
      for(int i = 0; i<7; i++){
          arr2[i] = pow(2, arr1[i]);    //arr2 will be created.
          cout<<arr2[i]<<" ";           //Show arr2.
          }
}

answered Dec 27 '17 at 11:53

deepjyoti30

22
2

yes thank you, other answers already explained very good, but your answer is also valid, so thank you. – Dec 28 '17 at 07:14

c++ why dont print all elements when exponent

3 Answers3