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

尝试 char pop(Node** top) 并在 (*top) 上操作

Answer 2

请将顶部指针的引用发送到 pop 函数，如“char pop(Node **top) { }”并添加更改您的 else 块“top[0] = top[0]->next;” 而不是“top = top->next ;”。

Answer 3

您还需要释放 top 的当前位置...使用 free as

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 ;
    }

    free(ptr); 

==================================================== ================

称它为

int main(){
 Node front = NULL:

 // place your code of PUSH here.

 pop(&front); // will call the function **pop**

}

}

Answer 4

我们需要更多代码，但我会尝试：

我猜你像这样使用这个函数：

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，然后stack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;
}

Answer 5

如果您更改变量的值，则将指针（其地址）传递给函数，例如

void increment(int *p) {
      p += 1;
}

类似于更改 POINTER 的值，您需要将指针传递给指向函数的指针

char pop(Node **top) {
       char t;
       Node  *p;
    if( size == 0 ) {
        return '\0';
    } else {
        p = *top;
        t = p->op;
        (*top) = (*top)-> next;
        free(p);
        return t;
    }
}