Why Does `changeValue(value)` Not Change `value` in C/C ?

The Conundrum of Call-by-Value: Unraveling the Value Preservation Puzzle

In C/C , function parameters are inherently passed by value. This implies that a replica of the original variable is passed to the function, rather than the original variable itself. This behavior can lead to confusion, as illustrated by the following code snippet:

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

int main() {
  int value = 5;
  changeValue(value);

  cout <p>Upon executing this code, one might expect the output to be "The value is: 6." after the function call. However, surprisingly, the output remains "The value is: 5." The reason for this behavior lies in the intricacies of call-by-value.</p><p>When the function changeValue() is invoked, a copy of the value 5 is created and passed to the function. This means that within the function, the value 6 is assigned to the copy, leaving the original value of 5 untouched.</p><p>To rectify this issue and modify the original variable, one must resort to call-by-reference. This technique involves passing a reference to the original variable to the function. In this case, the function can then modify the original value, reflecting the changes in the calling function.</p><pre class="brush:php;toolbar:false">void changeValue(int& value) {
  value = 6;
}

int main() {
  int value = 5;
  changeValue(value);

  cout <p>Now, upon executing the code, the output will be "The value is: 6." This highlights the distinction between call-by-value and call-by-reference and the importance of choosing the appropriate one based on the intended behavior of the program.</p>

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