如何对某个函数返回的最后一个节点执行操作？

Question

我在下面有一个双向链表。考虑两个函数some_function1 和some_function2。它们都返回指向某个节点的指针。现在，还有另一个函数do_something，它要么做一些事情(if some_condition = 1)，要么什么都不做(if some_condition = 0)。如果some_condition = 1，那么我希望它对some_function1 或some_function2返回的最后一个节点进行一些操作，以主函数最后一次调用的为准.如何做到这一点？谢谢

编辑：我真正想做的是删除 some_function1 或 some_function2 返回的最后一个节点。

typedef struct Node {

    int item; 
    struct Node *next; 
    struct Node *prev; 

} Node;

typedef struct List {

    Node *current_node;  

} List;



int some_condition = 0; 



int some_function1(List L){
    if (something){
        //do stuff
        some_condition = 1; 
    }
    else{
        //do stuff
        some_condition = 0; 
    }

    return *pointer to some node
}

int some_function2(List L){
    if(something){
        //do stuff
        some_condition = 1;
    }
    else{
        //do stuff
        some_condition = 0; 
    }

    return *pointer to some node 
}

//returns 1 or 0 based on success
int do_something(List L){
    if(some_condition == 1){
        //HERE, I WANT TO DO SOMETHING ON THE NODE THAT WAS LAST RETURNED BY EITHER some_function1 OR some_function2 
        return 1; 

    }
    else{
        return 0; 
    }
    some_condition = 0; 
}

Answer 1

如果我的理解正确，您希望do_something 函数使用some_function1 或some_function2 这两个函数返回的最近使用 节点执行操作。

您可以使用具有模块范围的static 指针来实现此目的，以缓存一个值：

static struct Node * MRU_node;

int some_function1(List L){
    if (something){
        //do stuff
        some_condition = 1; 
    }
    else{
        //do stuff
        some_condition = 0; 
    }

    MRU_node = *pointer to some node    // HERE
    return MRU_node;
}

然后在do_something函数中添加代码来引用这个缓存：

//returns 1 or 0 based on success
int do_something(List L){
    if (some_condition == 1){
        ... MRU_node ... ; // reference the cached pointer
        return 1; 

    }
    else{
        return 0; 
    }
    some_condition = 0; 
}