Squaring numbers in an array

Question

I want to know if there is a better way to square numbers in an array. This is what I wrote:

#include <iostream>
using std::cout;
using std::endl;


void square(int []);


int main()
{
    int array[] = {2,4,6,8,10};

    square(array);
    for(auto i : array)
        cout << i << " ";

}



void square(int array[])
{
    int size = sizeof(array)/sizeof(array[0]);
    for(int i=0; i< size; i++)
        array[i] *= array[i];
}

Answer 1

1. int size = sizeof(array)/sizeof(array[0]);

   Just so you know, sizeof() returns size_t.

2. You can use a vector instead of an array.

   vector<int> data = {2,4,6,8,10};
   

3. Simplify squaring using std::transform() and use a lambda expression in place of a function object:

#include <iostream>
#include <algorithm>
#include <functional>

using std::cout;
using std::transform;
using std::vector;

int main()
{

    vector<int> data = {2,4,6,8,10};

    transform(data.begin(), data.end(), data.begin(), [](int x){return x*x;});


    for(auto i : data)
        cout << i << " ";

}

Answer 2

(This looks like a homework question.)

void square(int array[])
{
    int size = sizeof(array)/sizeof(array[0]);

Here you have a major bug, which you would have noticed if you had compiled and run your code before posting it. Because array is a function parameter of type int*, sizeof(array) is the same as sizeof(int*). Therefore you square only the first 1 or 2 elements of your input array.

The most common idioms for passing around arrays in C++ are

void square(int *array, size_t length);  // a.k.a. "the C way"
void square(std::vector<int>& array);    // a.k.a. "dodging the question"
void square(int *begin, int *end);       // the STL does this, but template-ized
void square(int (&array)[5]);            // Pass an array by reference (not very useful as you need to specify the size) 
template<int N>
void square(int (&array)[N]);            // Use a template to make passing an
                                         // array by reference useful.
                                         // Note: You now no longer need to calculate the size (its N)

I guess my favorite "too advanced" answer for this one would be

#include <iostream>

template<class Array>
void square(Array& array) {
    for (auto& elt : array) {
        elt *= elt;
    }
}

int main() {
    int array[] = {2,4,6,8,10};
    square(array);
    for (auto i : array) {
        std::cout << i << " ";
    }
    std::cout << std::endl;
}