# C語言筆記

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這份筆記僅用於紀錄[你所不知道的 C 語言: linked list 和非連續記憶體](https://hackmd.io/@sysprog/c-linked-list)課堂中所學之內容

struct ListNode *mergeTwoLists(struct ListNode *L1, struct ListNode *L2) {
    struct ListNode *head = NULL, **ptr = &head, **node;

    for (node = NULL; L1 && L2; *node = (*node)->next) {
        node = (L1->val < L2->val) ? &L1: &L2;
        *ptr = *node;
        ptr = &(*ptr)->next;
    }
    *ptr = (struct ListNode *)((uintptr_t) L1 | (uintptr_t) L2);
    return head;
}

最後一行的code
`*ptr = (struct ListNode *)((uintptr_t) L1 | (uintptr_t) L2);`

可以參考[這是什麼黑魔法](https://facebook.com/hashtag/這是什麼黑魔法/?__gid__=1957891421058925)
重點就在於L1和L2其中一個指標已經指向NULL，透過uintptr_t的轉型，就可以將指向NULL的指標位址轉換為0

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以下簡單測試一下上述的狀況

#include <stdio.h>
#include <stdint.h>

typedef struct ListNode {
    int val;
    struct ListNode *next;
}ListNode;

int main() {
    ListNode node1 = {1, NULL};

    ListNode *L1 =  &node1;
    ListNode *head=NULL, *ptr=NULL;


    printf("(uintptr_t) L1: %p \n", (void*)(uintptr_t) L1);

    printf("(ListNode *)(uintptr_t) head: %p\n", (ListNode *)(uintptr_t) head);
    printf("(ListNode *)(uintptr_t) L1: %p\n", (ListNode *)(uintptr_t) L1);

    ptr = (struct ListNode *)((uintptr_t) L1 | (uintptr_t) head);
    printf("ptr: %p\n", (ptr));

    return 0;
}

ouput:

(uintptr_t) L1: 0x7fffffffdbd0 
(ListNode *)(uintptr_t) head: (nil)
(ListNode *)(uintptr_t) L1: 0x7fffffffdbd0
ptr: 0x7fffffffdbd0

最後ptr也還是和L1一樣指向node1