CODEWITHSUNDEEP
c++arrayspointers
zZzZ
zZzZ

Reputation: 173

Swapping the content of an array using pointer

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This is the sample output of the question.My swapping part is not functioning well...

int *smallest, *current = array;
    for(int i=0; i<MAX-1; i++, current++){ //partial code given by the question

        smallest = current;
        cout << "\nCurrent=" << *current;
        int tmp = *current;
        for(int j = i; j < MAX; j++,smallest++){ //my code 
            if(*smallest < tmp){
                tmp = *smallest;
            }
        }
        *smallest = tmp;
        cout << " Smallest=" <<*smallest<< endl;

        int tmp2; //not functioning from this line onwards
        tmp2 = *smallest;
        *smallest = *current;
        *current = tmp2;
     }

Upvotes: 0

Views: 220

Answers (3)

darune
darune

Reputation: 10962

First of all, you certainly do not need 2 for loops for this. That being said you may just the algorithm library directly for this task:

https://en.cppreference.com/w/cpp/algorithm/min_element

#include <vector>
#include <algorithm>

int main(){
    std::vector elements = {8,6,4,2,16};
    return *std::min_element(std::begin(elements), std::end(elements));
}

Try it yourself: https://godbolt.org/z/oHkMid

And using raw pointers you just need to do that instead - something like the following:

return *std::min_element(array, array + MAX);

or if array has known size

return *std::min_element(std::begin(array), std::end(array));

And what's wrong with you original code ? when you get to the last iteration of the outer loop you only examine minimum of the last element (main problem was: you have two loops).

Upvotes: 3

srt1104
srt1104

Reputation: 959

You don't need two for loops to find the smallest element in an array. It can be done in only one traversal of the array.

int *smallest, *current;
smallest = current = array;                     // both smallest and current refer to &array[0]
for (; current - array <= MAX - 1; ++current)   // current will increment till the last element of the array which is at MAX-1 distance from &array[0]
    if (*current < *smallest)                   // if element at *current is less than element at *smallest then update smallest.
        smallest = current;
current = nullptr;                              // current refers to the element at MAX distance from &array[0] hence it should be set to nullptr.

cout << *smallest << endl;

Upvotes: 2

4386427
4386427

Reputation: 44274

I'm don't fully understand your algorithm. If you just want to find the smallest element in an array, you don't need two for loops.

However, I do see some problems in your code.

for(int j = i ;j < MAX-i ; smallest++){ // Here you don't want to increment smallest !
                                        // You only want to change smallest
                                        // when *current is less than *smallest
                                        // You probably want ++j instead
                                        //
                                        // Further "MAX-i" should be "MAX-1"

   if(array[j] < *current){             // Here you don't want to compare array and *current
                                        // Instead you want:
                                        //     array[j] < *smallest

       smallest = &array[i];            // Here you want to use j instead of i:
                                        //      smallest = &array[j]
   }
}

If you "just" want to find the smallest element in the array using pointers, you can do it like:

  int *smallest = array;
  int *current = array;
  for(int i=0; i<MAX-1; i++, current++){
    cout << "\nCurrent=" << *current;

    if(*current < *smallest){
      smallest = current;
    }

    cout << " Smallest=" <<*smallest<< endl;
  }

Upvotes: 1

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