我正在努力克服C中的问题，我无法弄清楚发生了什么。 我正在根据字母的频率对树进行排序（例如“cab”为“1a 1b 1c”）。这是我的代码：
#define M 8 //maximal number of appearances of a letter
#define LAST_LETTER 25 //number of last letter
#define bzero(b,len) (memset((b), '\0', (len)), (void) 0)
static const char* alphabet = {"abcdefghijklmnopqrstuvwxyz"};
//TODO: define list structures
typedef struct word word;
struct word {
char* _word;
word *next;
};
typedef struct list list;
struct list {
word *first;
};
typedef struct dict dict;
struct dict {
dict *children[M];
list *words[M];
};
//returns an empty list
list * list_new() {
list *l = (list*) malloc(sizeof(list));
word *w = (word*) malloc(sizeof(word));
if (l == NULL || w == NULL) {
printf("could not create list : memory allocation failed\n");
return;
}
w->_word = NULL;
w->next = NULL;
l->first = w;
return l;
}
//append word at end of list
void list_append(list *l, char *w) {
// create a new word
word *new_word = malloc(sizeof(word));
if (l == NULL || new_word == NULL) {
printf("could not append word to list : list is empty\n");
return;
}
new_word->_word = malloc(strlen(w) + 1);
strcpy(new_word->_word, w);
new_word->next = NULL;
//insert the word
if (l->first->_word == NULL) {
l->first->_word = new_word->_word;
}
else {
//word *temp = malloc(sizeof(word));
word *temp;
temp = l->first;
while(temp->next != NULL) {
temp=temp->next;
}
temp->next = new_word;
}
}
//print word list
void list_print(list *l) {
if (l == NULL || l->first == NULL) {
printf("could not print list : list is empty\n");
return;
}
word *current = l->first;
while (current != NULL) {
printf("%s -> ", current->_word);
current = current->next;
}
printf("NULL\n");
}
char *compute_signature(const char *word) {
char *signature = (char*) malloc(26);
memset((void*)signature, 0, 26);
int i = 0, j = 0, n = 0;
char current_letter, letter;
for (i = 0; i < 26; i++) {
current_letter = alphabet[i];
n = 0;
for (j = 0; j < (int) strlen(word); j++) {
letter = word[j];
if (letter == current_letter) {
n++;
}
}
signature[i] = (char) n;
}
return signature;
}
void dict_insert(dict *d, char *signature, unsigned int current_letter, char *w) {
int j = 0;
int steps = 0;
int occur = 0;
dict *temp = NULL;
if (current_letter == strlen(w)-1) {
printf("Word found : %s!\n",w);
int i = 0;
int different_letters = 0;
for (i = 0; i < 26; i++) {
if ((int) signature[i] > 0) {
different_letters++;
}
}
for (i = 0; i < 26; i++) {
occur = (int) signature[i];
if (occur > 0) {
steps++;
if (steps < different_letters) break;
}
else {
if (temp == NULL) {
temp = d;
}
if (temp->children[occur] == NULL) {
temp->children[occur] = (dict*) malloc(sizeof(dict));
temp = temp->children[occur];
}
}
}
if (temp == NULL) {
temp = d;
}
list *l = NULL;
if (temp->words[occur] == NULL || temp->words[occur]->first == NULL) {
temp->words[occur] = list_new();
l = temp->words[occur];
}
else {
l = temp->words[occur];
}
char *new;
new = malloc(strlen(w) + 1);
strcpy(new, w);
list_print(l);
/*list_append(l,new);
list_print(l);*/
}
else {
printf("Current letter: %c.\n",w[current_letter]);
dict_insert(d,signature,current_letter+1,w);
}
}
dict * read_words(const char *file) {
FILE* f = NULL;
f = fopen(file, "r");
if (f == NULL) {
printf("Could not open file.\n");
exit(EXIT_FAILURE);
}
char line[256];
dict *d = (dict*) malloc(sizeof(dict));
//bzero((void*)d, sizeof(dict));
while (fgets(line, sizeof(line), f) != NULL) {
if (line[strlen(line) - 1] == '\n') {
line[strlen(line) - 1] = '\0';
}
char *new;
new = malloc(strlen(line) + 1);
strcpy(new, line);
char *signature = compute_signature(new);
dict_insert(d, signature, 0, new);
free((void*)signature);
}
return d;
}
int main(int argc, const char* argv[]) {
/*list *myList = list_new();
list_print(myList);
list_append(myList,"Word1");
list_print(myList);
list_append(myList,"Word2");
list_print(myList);
list_append(myList,"Word3");
list_print(myList);*/
dict *d = read_words("list");
/*list_print(d->words[1]);
list_print(d->children[0]->words[1]);
list_print(d->children[0]->children[0]->words[1]);
list_print(d->words[2]);
list_print(d->children[0]->words[2]);*/
return 0;
}
此代码几乎可以使用。但不是按照预期的：有些单词在树中放错了位置，有时列表似乎是重复的（我还没有弄清楚到底是怎么回事）。你能帮我清理一下这段代码吗（也许可以指出最大的错误^^）？
修改：部分修复了代码的代码。我在list_append()上的if (strcmp(l->first->_word, "") == 0)处有一个段错误。
编辑2 ：当我在算法中传递一个单词时，我需要测试列表是否存在：例如我将“厄运”添加到对应于“1d 1m 2o”的列表中，当我传递“情绪”时，我希望算法将其添加到同一列表中。以下是我试图实现这一目标的方法：
if (temp == NULL) {
temp = d;
}
list *l = NULL;
if (temp->words[occur] == NULL || temp->words[occur]->first == NULL) {
temp->words[occur] = list_new();
l = temp->words[occur];
}
else {
l = temp->words[occur];
}
我有temp->words[occur]->first == NULL ...
编辑3 ：我的代码正在编译，但这些字词未添加到右侧列表中。我认为问题出在dict_insert()，声明是：
void dict_insert(dict *d, char *signature, unsigned int current_letter, char *w) {
int j = 0;
int steps = 0;
int occur = 0;
dict *temp = NULL;
if (current_letter == strlen(w)-1) {
printf("Word found : %s!\n",w);
int i = 0;
int different_letters = 0;
for (i = 0; i < 26; i++) {
if ((int) signature[i] > 0) {
different_letters++;
}
}
for (i = 0; i < 26; i++) {
occur = (int) signature[i];
if (occur > 0) {
steps++;
if (steps == different_letters) break;
}
printf("%d RIGHT\n",occur);
printf("1 DOWN\n");
if (temp == NULL) {
temp = d;
}
temp->children[occur] = (dict*) realloc(temp->children[occur], sizeof(dict));
//temp = temp->children[occur]
}
printf("%d RIGHT\n",occur);
if (temp == NULL) {
temp = d;
}
if (temp->words[occur] == NULL) {
list *l = list_new();
temp->words[occur] = l;
}
char *new;
new = malloc(strlen(w) + 1);
strcpy(new, w);
list_print(temp->words[occur]);
list_append(temp->words[occur],new);
list_print(temp->words[occur]);
}
else {
printf("Current letter: %c.\n",w[current_letter]);
dict_insert(d,signature,current_letter+1,w);
}
}
编辑4 ：dict_insert()完全重做。
void dict_insert(dict *d, char *signature, unsigned int current_letter, char *w) {
int occur;
occur = (int) signature[current_letter];
if (current_letter == LAST_LETTER) {
if (d->words[occur] == NULL) {
d->words[occur] = list_new();
}
char *new;
new = malloc(strlen(w) + 1);
strcpy(new, w);
printf("word found : %s!\n",w);
list_print(d->words[occur]);
list_append(d->words[occur],new);
list_print(d->words[occur]);
}
else {
if (d->children[occur] == NULL) {
d->children[occur] = malloc(sizeof(dict));
}
d = d->children[occur];
dict_insert(d,signature,current_letter+1,w);
}
}
仍然是if (d->children[occur] == NULL) ...
答案 0 :(得分：3)
在compute_signature中，您试图将字母表的当前字母与单词的当前字母进行比较
strcmp((const char*) &letter, (const char*) ¤t_letter) == 0
letter和current_letter为char *。因此，您通过letter = alphabet[i]将它们指向内存，然后获取指针的地址并将其粘贴到strcmp中。我很惊讶它没有在原版中崩溃。
letter和current_letter应更改为char而不是char *，您的比较应为if (letter == current_letter)
其他一些评论：
malloc的返回值。虽然这不一定会打破任何事情not a good idea。int和char之间进行投射。虽然您不可能在同一个字母上出现超过127个单词，但您应该考虑将签名更改为short或int 在list_append进行
word *temp = malloc(sizeof(word));
temp = l->first;
这会在您分配一些内存时产生内存泄漏，然后通过将temp设置为列表头来忘记它。只是摆脱malloc没有必要。 dict_insert
temp->words[occur] = (list*) malloc(sizeof(list));
list *l = list_new();
temp->words[occur] = l;
摆脱malloc。