c - 初始化结构指针数组的正确方法是什么？

Question

我正在尝试实现 trie 数据结构:-

typedef struct tries{
    char university[20];
    struct tries *path[10];
} tries;

tries* head = (tries*)malloc(sizeof(tries));
head->path = { NULL } ;

每当我尝试将路径数组的所有元素初始化为 NULL 时，我都会收到此错误:-

clang -fsanitize=signed-integer-overflow -fsanitize=undefined -ggdb3 -O0 -std=c11 -Wall -Werror -Wextra -Wno-sign-compare -Wshadow    tries.c  -lcrypt -lcs50 -lm -o tries
tries.c:20:18: error: expected expression
    head->path = { NULL } ;
                 ^
1 error generated.
make: *** [tries] Error 1

如何将所有路径数组的元素初始化为 NULL
我在我的插入和搜索函数中使用了这个 NULL 值。

void Search(char* university, char* year, tries* head){
    int pos = 0;
    int length = strlen(year);
    tries* temp = head;

    for(int i = 0 ; i < length ; i++){
        pos = (int) (year[i]%48);

        if(temp->path[pos] != NULL){
            temp = temp->path[pos];
        } else {
            printf("%s Not Found !!\n", university);
            return;
        }

    }

    if(strcmp(temp->university, university) == 0){
        printf("%s is Presnt.\n", university);
    } else {
        printf("%s Not Found !\n", university);
    }

}

Answer 1

只需使用 calloc()而不是 malloc()你会得到所有0 s 在 char -数组和所有NULL s 进入指针数组。

改变

tries * head = (tries*)malloc(sizeof(tries));

成为

tries * head = (tries*)calloc(1, sizeof(tries));

另请注意

• 不需要投void -C 中的指针
• sizeof是运算符不是函数

就这样吧:

tries * head = calloc(1, sizeof (tries));

如果您希望此代码行更健壮，可以在类型更改后幸存下来 head做到这一点

tries * head = calloc(1, sizeof *head);

---

当你处理 struct 时实际上是可分配的 s 您可以执行以下操作:

const tries init_try = {0};

...

  tries * head = malloc(sizeof *head);
  if (NULL == head)
    exit(EXIT_FAILURE);
  *head = init_try;