How Can I Modify a Pointer Passed to a C Function and Reflect Those Changes in the Calling Scope?

Function Behavior: Passing Pointers in C

In C , passing a pointer to a function allows for modifying the original variable in the calling scope. However, certain nuances can arise when dealing with pointers.

Issue:

Consider a function that modifies a passed pointer, as in the following example:

bool clickOnBubble(sf::Vector2i &mousePos, std::vector<Bubble *> bubbles, Bubble *targetBubble) {
    targetBubble = bubbles[i];
}

Despite modifying the 'targetBubble' pointer within the function, changes are not reflected outside of it.

Reason and Solution:

The issue stems from passing a copy of the pointer rather than a reference to it. To rectify this, you can utilize the following techniques:

1. Pointer to Pointer:
   Pass a pointer to the pointer, which allows you to modify the original pointer:

   void foo(int **ptr) //pointer to pointer
   {
       *ptr = new int[10]; //just for example, use RAII in a real world
   }

2. Reference to Pointer:
   Pass a reference to the pointer, which directly modifies the original pointer:

   void bar(int *&ptr) //reference to pointer (a bit confusing look)
   {
       ptr = new int[10];
   }

By implementing either approach, you can effectively modify the pointer in the calling scope, ensuring that changes made within the function are reflected externally.

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