c语言 - 哈希表的键由于上次输入而混乱

Question

我正在尝试创建一个哈希表程序，该程序可以输入哈希键和将存储在哈希表中的字符串数据。一切都很好，除了在"菜单==2"中，每次我试图通过输入哈希键来搜索字符串数据时，表中的一个(最后一个定位的(哈希键都会更改为我在"菜单===2"中输入的哈希键。

这是来自"Menu==4"的带有viewAll ()功能的哈希表(在我通过"Menu==1"输入哈希键和字符串数据之后(：

//FORMAT:
//[index]: [key] (string data) -> [key] (string data) -> ..
--------------------------------------------------------------------------------
//after I finished inputting all hash keys and string datas through menu 1  
[0]: 0 (INDIA) -> 97 (PAKISTAN) -> 194 (BURMA) -> 291 (AFGHANISTAN)
[1]: 1 (UNITED STATES) -> 98 (CANADA) -> 195 (MEXICO) -> 292 (PANAMA)
[2]: 2 (THE NETHERLANDS) -> 99 (LUXEMBOURG) -> 196 (BELGIUM) -> 293 (FRANCE)
[3]: 3 (SYRIA) -> 100 (LEBANON) -> 197 (ISRAEL) -> 294 (BAHRAIN)
//this hash table is correct
--------------------------------------------------------------------------------
after menu 2 (last key inputted on menu 2 was 0)
[0]: 0 (INDIA) -> 97 (PAKISTAN) -> 194 (BURMA) -> 291 (AFGHANISTAN)
[1]: 1 (UNITED STATES) -> 98 (CANADA) -> 195 (MEXICO) -> 292 (PANAMA)
[2]: 2 (THE NETHERLANDS) -> 99 (LUXEMBOURG) -> 196 (BELGIUM) -> 293 (FRANCE)
[3]: 3 (SYRIA) -> 100 (LEBANON) -> 197 (ISRAEL) -> 0 (BAHRAIN)
//(why is BAHRAIN key 0, it should be 294)
--------------------------------------------------------------------------------
after menu 2 (last key inputted on menu 2 was 99)
[0]: 0 (INDIA) -> 97 (PAKISTAN) -> 194 (BURMA) -> 291 (AFGHANISTAN)
[1]: 1 (UNITED STATES) -> 98 (CANADA) -> 195 (MEXICO) -> 292 (PANAMA)
[2]: 2 (THE NETHERLANDS) -> 99 (LUXEMBOURG) -> 196 (BELGIUM) -> 293 (FRANCE)
[3]: 3 (SYRIA) -> 100 (LEBANON) -> 197 (ISRAEL) -> 99 (BAHRAIN)
//(why is BAHRAIN key 99, it should be 294)

正如您所看到的，它总是最后一个被输入菜单2更改的位置。

下面是源代码，为了不浪费时间，我会看看void viewStrData ()函数。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
struct data {
int key;
char strData [100];
struct data *next;
}*head [97], *tail [97], *node, *curr;
//index location within hash table = key % 97
int hashing (int key) {
int result = key % 97;
return result;
}
//input key and strData, push it to hash table
void push (int key, char strData [100]) {
node = (struct data*) malloc(sizeof(struct data));
node->key = key;
strcpy (node->strData, strData);
node->next = NULL;
int idx = hashing (key);
if (!head [idx]) {
head [idx] = tail [idx] = node;
} else {
tail [idx]->next = node;
tail [idx] = node;
}
}
//see strData based on key 
void viewStrData (int key) {
curr = head [hashing (key)];
node->key = key;
while (true) {
if (curr->key == node->key) {
printf ("[%s]n", curr->strData);
break;
} else {
curr = curr->next;
}
}
}
//print out current hash table
void viewAll () {
for (int i = 0; i < 97; i++) {
printf ("[%d]: ", i);
if (!head [i]) {
printf ("-n");
} else {
curr = head [i];
while (curr) {
if (curr == head [i]) {
printf ("%d", curr->key);
printf (" (%s)", curr->strData);
} else {
printf (" -> %d", curr->key);
printf (" (%s)", curr->strData);
}
curr = curr->next;
}
printf ("n");
}
}
}
int main () {
int menu, key;
char strData [100];
while (true) {
printf ("=======================================n");
printf ("MENU [1-3]: n");
printf ("1. INPUT DATAn");
printf ("2. SEARCH DATAn");
printf ("3. EXITn");
printf ("=======================================n");
scanf ("%d", &menu);
if (menu == 1) {
printf ("ENTER KEY AND STRING DATA: n");
scanf ("%d %[^n]", &key, strData);
push (key, strData);
printf ("HASH TABLE UPDATED!n");
} else if (menu == 2) {
printf ("ENTER KEY TO FIND STRING DATA: n");
scanf ("%d", &key);
printf ("STRING DATA OF KEY [%d] IS: ", key);
viewStrData (key);
} else if (menu == 3) {
printf ("PROGRAM EXITEDn");
break;
} else if (menu = 4) {
viewAll ();
}
}
return 0;
}

有什么办法我可以解决这个问题吗？我事先很抱歉，我的代码读起来很草率和糟糕。

Answer 1

之所以会发生这种情况，是因为每次调用push时，都会分配新的内存，并将指向该内存的指针存储在节点中：

node = (struct data*) malloc(sizeof(struct data));

之后，该分配的数据被插入到表中，但节点仍然指向该数据。然后，当您运行viewStrData时，您执行node->key=key，此时节点仍然指向您调用push的最后插入的数据。这就是为什么每次调用viewStrData时，最后插入的节点的密钥都会更新为接收到的密钥。

我建议直接比较收到的密钥，如下所示：

void viewStrData (int key) {
curr = head [hashing (key)];
// node->key = key;   <-- No need for this
while (true) {
if (curr->key == key) { // <-- compare key directly, instead of storing it into node->key
printf ("[%s]n", curr->strData);
break;
} else {
curr = curr->next;
}
}

附言：这行有个打字错误：

} else if (menu = 4) {

它应该是=而不是=。当前，如果菜单不是1、2或3，它将在if语句中为其分配4，并且每次4的结果都将true。因此，即使您在那里输入7，菜单也将被4覆盖，并且将调用viewAll。