c - 如何通过引用传递结构并更改其值

Question

大家好，我的 C 代码有这个问题。我正在实现一个堆栈，每当我弹出它时，它都会更改函数 pop 中的堆栈，但不会更改原始堆栈。请帮忙。

这是我的代码

char pop(Node* top)
{
    char result ; 

    if (size == 0) // The stack is empty.
    {
        return '\0' ;
    }

    else //The stack contains at least one element .
    {
        result = top->opp ;
        top = top->next ;
    }

    size-- ;

    return result;
}

Answer 1

我们需要更多代码，但我会尝试:

我猜你是这样使用这个函数的:

char pop(Node* top)
{
    char result ; 

    if (size == 0) // The stack is empty.
    {
        return '\0' ;
    }

    else //The stack contains at least one element .
    {
        result = top->opp ;
        top = top->next ;
    }

    size-- ;

    return result;
}

int main()
{
   // this is the top of the stack
   Node    *stack; // let's say there already are some elements in this stack
   pop(stack);
   return 0;
}

问题是你要改变指针的值，那么stack就会指向栈顶。为此，您必须使用指向指针的指针，如下所示:

char pop(Node** top)
{
    char result ; 
    Node *ptr;

    ptr = *top;
    if (size == 0) // The stack is empty.
    {
        return '\0' ;
    }

    else //The stack contains at least one element .
    {
        result = (*top)->opp ;
        (*top) = (*top)->next ;
        // You should free the pointer, like user2320537 said in his answer.
        free(ptr);
    }

    size-- ;

    return result;
}

int main()
{
   // this is the top of the stack
   Node    *stack; // let's say there already are some elements in this stack
   pop(&stack); // You give the pointer address
   return 0;
}