Why Can't I Modify Input Parameters in C/C Functions Using Call-by-Value?

Why Call-by-Value Fails to Update Input Parameters

In C/C , function parameters are typically passed by value, meaning that a copy of the input variable is created and sent to the function. This implies that any modifications made within the function only affect the local copy and not the original variable.

Consider the example:

#include <iostream>

using namespace std;

void changeValue(int value);

int main() {
  int value = 5;
  changeValue(value);
  cout << "The value is: " << value << "." << endl;
  return 0;
}

void changeValue(int value) {
  value = 6;
}

This code attempts to modify the value of the input parameter within the changeValue function. However, the output remains 5, even though the function changes the local copy to 6.

This is because the changeValue function receives a copy of the value variable from main(). The function then operates on this local copy, which is independent of the original variable in main(). Therefore, the changes made within the function do not carry over to the original variable.

Using Call-by-Reference to Modify Input Parameters

To modify the original variable from within a function, call-by-reference must be used. This allows the function to access and modify the actual variable in memory, rather than just a copy:

#include <iostream>

using namespace std;

void changeValue(int &value);

int main() {
  int value = 5;
  changeValue(value);
  cout << "The value is: " << value << "." << endl;
  return 0;
}

void changeValue(int &value) {
  value = 6;
}

In this example, the changeValue function takes a reference to the value variable, represented by the ampersand (&). This allows the function to access and modify the original variable directly. As a result, when the function sets the value to 6, it modifies the actual variable in main(), and the output correctly reflects the change.

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