Task: you need to find common elements that are contained in two arrays at the same time. At the output, the program should produce information of the form: {a, b, c, ..., n}, where a, b, c, n are the elements of the array. I only get (if I use the code that is commented out) to make an output of the type: {a, b, c,}, which is an error, since I give the labs through the site (progtest, if anyone knows).

Therefore, I see a new solution: I create another array in which I will store the same values ​​of the arrays of the first two. And if you print the values ​​of the array at that moment, when the values ​​are saved in the array, then everything is ok. When I want to display the values ​​of the array already in a separate cycle, following after the cycle of finding the elements, it turns out porridge.

Example:

arr1 = 1, 2, 3, 4, 5

arr2 = 2, 3, 4, 7, 8, 9

output of array data immediately after storing values ​​in it: 2, 3, 4

output array data in a separate cycle: 5838432 2 3 4 83886085 4822

#include <iostream> using namespace std; int main() { int size1, size2; int len1, len2; cout << "Enter size of set A:\n"; cin >> size1; if (cin.fail() || size1 < 0) { cout << "Invalid input.\n"; return 0; } int arr1[size1]; len1 = sizeof(arr1) / sizeof(*arr1); cout << "Enter members of set A:\n"; for (int i = 0; i < len1; i++) cin >> arr1[i]; if (cin.fail()) { cout << "Invalid input.\n"; return 0; } cout << "Enter size of set B:\n"; cin >> size2; if (cin.fail() || size2 < 0) { cout << "Invalid input.\n"; return 0; } int arr2[size2]; len2 = sizeof(arr2) / sizeof(*arr2); cout << "Enter members of set B:\n"; for (int i = 0; i < len2; i++) cin >> arr2[i]; if (cin.fail()) { cout << "Invalid input.\n"; return 0; } int* intersect = new int[len2]; for (int i = 0; i < len1 ; i++) { for (int j = 0; j < len2; j++) { if (arr1[i] == arr2[j]) { //cout << arr1[i] << ", "; intersect[i] = arr1[i]; cout << intersect[i] << " "; } } } cout << endl; for (int i = 0; i < len2; i++) { cout << intersect[i] << " "; } return 0; }
  • If the elements are duplicated, for example, {1, 1, 2} and {1, 1, 1}, what should be displayed? - Vlad from Moscow
  • In this case {1} should be displayed - Vladislav Starikov
  • If I know the pluses badly, I will be sincerely glad if you point out mistakes and tell me how to fix it - Vladislav Starikov
  • one
    I did not understand, and what you have is a mistake, as you wrote in the message? - Vlad from Moscow
  • The error is the wrong output format. That is, the site where I rent this lab, writes that the conclusion is wrong. And, of course, I don’t understand the rule for filling this “trash” into the array - Vladislav Starikov

1 answer 1

Your main mistake is that you do not assign intersection elements in a row:

 intersect[i] = arr1[i];

So if, say, a zero element in arr1 does not have the same in arr2 , there will be garbage in the zero element intersect .

There are others. I corrected your code, look. But keep in mind - it will not work if the sets are in fact multisets in which there may be duplicate elements . I didn’t fix your algorithm here, I’m just warning you about such troubles - think for yourself how to deal with this.

 #include <iostream> using namespace std; int main(){ int size1, size2; cout << "Enter size of set A:\n"; cin >> size1; if(cin.fail() || size1<0){ cout << "Invalid input.\n"; return 0; } int * arr1 = new int[size1]; cout << "Enter members of set A:\n"; for(int i = 0; i < size1; i++) cin >> arr1[i]; if(cin.fail()){ cout << "Invalid input.\n"; return 0; } cout << "Enter size of set B:\n"; cin >> size2; if(cin.fail() || size2<0){ cout << "Invalid input.\n"; return 0; } int * arr2 = new int[size2]; cout << "Enter members of set B:\n"; for(int i = 0; i<size2; i++) cin >> arr2[i]; if(cin.fail()){ cout << "Invalid input.\n"; return 0; } int ilen = (size1 < size2) ? size1 : size2; int *intersect = new int[ilen]; ilen = 0; for (int i=0; i < size1 ; i++){ for (int j=0; j < size2; j++){ if (arr1[i] == arr2[j]){ //cout << arr1[i] << ", "; intersect[ilen] = arr1[i]; cout << intersect[ilen] << " "; ++ilen; } } } cout << endl; for(int i = 0; i< ilen; i++){ cout << intersect[i] << " "; } delete[] arr1; delete[] arr2; delete[] intersect; return 0; }