Sort integer numbers in ascending order using pointers?

Answer 1

It's hardly the most efficient way to sort integers. Integers are primitive data types, so it's actually quicker to sort the integers directly rather than point at them. Pointers just add an unnecessary level of indirection. Pointers should only be used for complex data types for which copying/moving would result in greater inefficiency than would otherwise be incurred through pointer manipulation.

#include<iostream>

#include<vector>

int main()

{

// instantiate a vector of unsorted integers

std::vector<int> v = {7,3,5,2,9,4,1,6,8};

// instantiate a vector of pointers to the integers in the vector

std::vector<int*> p;

for (size_t i=0; i<v.size(); ++i)

p.push_back (&v[i]);

// sort the pointers (uses insertion sort algorithm):

for (size_t i=1; i<p.size(); ++i)

{

int* x=p[i];

size_t gap = i;

size_t pre = i-1;

while (gap && *x<*p[pre])

{

p[gap--] = p[pre--];

}

p[gap]=x;

}

// print to prove the sorting worked:

std::cout << "Original order:\t\t\t";

for (size_t i=0; i<v.size(); ++i)

std::cout << v[i] << '\t';

std::cout<<std::endl;

std::cout << "Sorted order (by pointer):\t";

for (size_t i=0; i<p.size(); ++i)

std::cout << *p[i] << '\t';

std::cout<<std::endl;

}