方法 HASH_ADD 不会将新元素添加到散列 table。使用 uthash.h

Question

根据找到的文档here，我在C:

中编写了以下代码

adj_hash_table.h

typedef struct {
    int id_0;
    int id_1;
}id_t_;

typedef struct {
    id_t_ id;
    double value;
    UT_hash_handle hh;
}cell_t;

void add_(int id_0, int id_1, double value, cell_t *cells);
void free_table( cell_t *cells);
void main();

adj_hash_table.c

#include <stdio.h>
#include "uthash/src/uthash.h"
#include "adj_hash_table.h"

void add_(int id_0, int id_1, double value, cell_t *cells){

    cell_t l, *p;

    memset(&l, 0, sizeof(cell_t));
    l.id.id_0 = id_0;
    l.id.id_1 = id_1;
    HASH_FIND(hh, cells, &l.id, sizeof(id_t_), p); 

    if (p == NULL) {
        printf("Not found %d, %d\n", id_0, id_1);
        p = (cell_t *)malloc(sizeof *p);
        memset(p, 0, sizeof *p);
        p->id.id_0 = id_0;
        p->id.id_1 = id_1;
        HASH_ADD(hh, cells, id, sizeof(id_t_), p);
    }
    else
    {
        printf("Found %d, %d\n", id_0, id_1);
    }
    p->value = value;
}

void free_table( cell_t *cells){
    cell_t *p, *tmp;
    HASH_ITER(hh, cells, p, tmp) {
        HASH_DEL(cells, p);
        free(p);
    }
}

void main(){
    int nb_cells;
    cell_t *cells = NULL; 

    add_(0,0,1.0,cells);
    add_(0,1,2.0,cells);
    add_(0,0,3.0,cells);

    nb_cells=HASH_COUNT(cells);
    printf("number of cells: %d\n", nb_cells);

    free_table(cells);
}

当我使用 gcc -g -Wall -o adj_hash_table adj_hash_table.c 编译它并稍后使用 ./adj_hash_table 运行编译它时，我得到以下输出：

Not found 0, 0
Not found 0, 1
Not found 0, 0
number of cells: 0

但我预计：

Not found 0, 0
Not found 0, 1
Found 0, 0
number of cells: 2

这让我觉得 HASH_ADD 不起作用。 here 中的示例对我来说效果很好。我究竟做错了什么？另外，我的 free_table 方法是否正确？谢谢！！

Answer 1

来自参考文档的“将散列指针传递给函数”部分：

In the example above users is a global variable, but what if the caller wanted to pass the hash pointer into the add_user function? At first glance it would appear that you could simply pass users as an argument, but that won’t work right.

You really need to pass a pointer to the hash pointer:

The reason it’s necessary to deal with a pointer to the hash pointer is simple: the hash macros modify it (in other words, they modify the pointer itself not just what it points to).

也就是说，您需要将 cell_t ** 传递给您的 add_ 函数而不是 cell_t *，然后使用 *cells 调用 HASH 宏。

void add_(int id_0, int id_1, double value, cell_t **cells){
    ....
    HASH_FIND(hh, *cells, &l.id, sizeof(id_t_), p);
    ....
    HASH_ADD(hh, *cells, id, sizeof(id_t_), p);

调用将是：

cell_t *cells = NULL; 
add_(0,0,1.0,&cells);
add_(0,1,2.0,&cells);
add_(0,0,3.0,&cells);

方法 HASH_ADD 不会将新元素添加到散列 table。使用 uthash.h

The method HASH_ADD does not add the new elements to the hash table. Using uthash.h

c

hash

uthash