How is the pointer (*) making changes in the following code? - c++

Question

The following program is intended to check if a given element is in a given array, indices of array where the element occurs and number of times the element occurs. But, it doesn't give right results. I tried to replace poscount in seqsearch function with *poscount and did further changes for this pointer data type. Then the code works well. Why this is so?
#include <iostream>
using namespace std;
const int SIZE = 100;
void seqsearch(int[], int, int, int[], short);
int main() {
int array[SIZE], indices[SIZE];
int num, value;
short count = 0;
cerr << " Give number of elements in array : ";
cin >> num;
cerr << " Key in the array elements ";
for(int i = 0; i < num; i++) cin >> array[i];
cout << endl;
cerr << " Give the value to be searched : " << endl;
cin >> value;
cout << endl;
seqsearch(array, num, value, indices, count); // void function
if(count >= 0) {
cout << value << " found in array " << count << " times"
<< " at index positions " << endl;
for(int i = 0; i < count; i++) cout << indices[i] << " ";
cout << endl;
} else
cout << value << " not found in array " << endl;
return 0;
}
void seqsearch(int arr[], int size, int elm, int pos[], short poscount) {
int i, item;
poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[poscount] = i;
poscount = poscount + 1;
}
}
return;
}

Answer 1

The function seqsearch is supposed to return the result in pos and poscount, but the function takes poscount by-value which means that any changes you make to poscount inside the function, will be local to the function and not visible from the call site.
If you change the function to take the argument by-reference, the changes you make inside the function will actually be made to the variable used in the call to the function. Like this:
seqsearch(int arr[], int size, int elm, int pos[], short& poscount) // note short&
The int pos[] does not have the same problem because arrays decay into pointers, so it could have been int* pos instead - and that pointer points at the same array that you passed in at the call site.
Also note that the check after the call will make the program display "found in array" even if it isn't found in the array because the condition checks if count is zero or greater than zero.
if(count >= 0) { // should be if(count > 0) {
Suggestions unrelated to the problem in your question:
When the number of elements is not known at the time you compile your program, prefer to use a container which can grow dynamically, like a std::vector<int>. In your program you have a hardcoded limit of SIZE number of elements, but:
You will rarely use all of them.
You do not check if the user wants to enter more than SIZE elements and your program will gladly try to write out of bounds - which would cause undefined behavior.
Divide the program's subtasks into functions. It'll be easier to search for bugs if you can test each individual function separately.
Check that extracting values from std::cin actually succeeds.
int number;
if(std::cin >> number) { /* success */ } else { /* failure */ }
Check that the values entered makes sense too.
int wanted_container_elements;
if(std::cin >> wanted_container_elements && wanted_container_elements > 0) {
/* success */
} else {
/* failure */
}

Answer 2

poscount (or count in the context of the caller) in your code seems to be expected to be an output parameter.
To modify the passed value you must either have its address (a pointer) or a reference to the value.
Currently you are using "pass-by-value", meaning that poscount is a copy of count.
The original count stays untouched.
My personal favorite would be to return the value instead of using an out-parameter:
short seqsearch(int arr[], int size, int elm, int pos[]) {
int i, item;
short poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[poscount] = i;
poscount = poscount + 1;
}
}
return poscount;
}
count = seqsearch(array, num, value, indices);
Alternatively you can use a reference to manipulate the out-parameter:
void seqsearch(int arr[], int size, int elm, int pos[], short& poscount) {
int i, item;
poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[poscount] = i;
poscount = poscount + 1;
}
}
return;
}
seqsearch(array, num, value, indices, count);
And, as you already tried, you can also solve this by passing a pointer to the value:
void seqsearch(int arr[], int size, int elm, int pos[], short* poscount) {
int i, item;
*poscount = 0;
for(i = 0; i < size; i++) {
if(arr[i] == elm) {
pos[*poscount] = i;
*poscount = *poscount + 1;
}
}
return;
}
seqsearch(array, num, value, indices, &count);

Answer 3

When you pass your posscount argument, you pass a copy to the count variable in main, not the variable itself. That's why it works, when you pass it by pointer. You can also pass by reference. https://www.includehelp.com/cpp-tutorial/argument-passing-with-its-types.aspx