尝试释放链表时的 SIGABRT
SIGABRT while attempting to free a linked list
我正在研究我们的教授给我们准备即将到来的考试的一些旧课文,我 运行 解决了这个问题。
我的任务是从结构如下的文本文件中读取信息:
[十进制数],[罗马数(字符串)],[o或u(优化或未优化的罗马数)]
几千行并将该信息存储在二叉搜索树中,使用十进制数作为键。每个 b运行ch 还必须包含该数字出现的次数和遇到的各种罗马版本的列表,优化的版本位于列表的顶部。
然后释放一切。
我的代码(c):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct list //list node
{
char *rom;
struct list *next;
};
struct branch //main tree branch
{
int count, dec;
struct list *list;
struct branch *right, *left;
};
struct list *insertnode(struct list *head, char *rom, char opt) //creates a head or creates and adds a new node to the end
{
struct list *new;
if(!head) //if !head, make one
{
new = malloc(sizeof(new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = NULL;
return new;
}
if(opt == 'o') //if the roman form is optimized, put it in front of all the others
{
new = malloc(sizeof(new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = head;
return new;
}
head->next = insertnode(head->next, rom, opt); //recursive insertions
return head;
}
struct branch *insertbranch(struct branch *root, int dec, char *rom, char opt) //creates a root or creates and adds a new branch
{
struct branch *new;
if(!root) //if !root, make a root...
{
new = malloc(sizeof(new));
new->list = insertnode(new->list, rom, opt);
new->dec = dec;
new->count = 1;
new->right=new->left=NULL;
return new;
}
if(dec<root->dec) root->left = insertbranch(root->left, dec, rom, opt); //branch on the left, recursive
else if(dec>root->dec) root->right = insertbranch(root->right, dec, rom, opt); //branch on the right, recursive
else //if there already is such a branch, increase its count
{
root->count += 1;
root->list = insertnode(root->list, rom, opt);
}
return root;
}
void freelist(struct list *head) //frees list 'head'
{
struct list *tmp;
while(head)
{
tmp = head;
head = head->next;
free(tmp->rom);
free(tmp); // <- OFFENDING LINE
}
return;
}
void freetree(struct branch *root) //frees tree 'root'
{
if(!root) return;
freetree(root->left); //recursive!
freetree(root->right); //and on the right
free(root->right);
free(root->left);
freelist(root->list);
free(root);
return;
}
int main()
{
struct branch *root;
struct list *list;
int dec, i=1, n;
char rom[30], opt;
FILE *file = fopen("rom.csv", "r");
if(!file) //is the file even there?
return 1;
while(fscanf(file, "%d,%[^,],%c\n", &dec, rom, &opt)==3) //go through the file and fill tree
root = insertbranch(root, dec, rom, opt);
freetree(root);
printf("Goodbye!\n");
return 0;
}
调试后我确定了问题:在函数 "freelist" 中,当它到达命令 "free(tmp)" 时程序中止。我不知道可能是什么原因。我什至检查以确保节点头存在。
感谢您的帮助!
您没有在 insertnode()
和 insertbranch()
中分配正确的内存量。
你需要替换这个:
new = malloc(sizeof(new));
有:
new = malloc(sizeof(*new));
这是因为new
是一个指针。使用 malloc(sizeof(new))
,您只分配 space 来存储指针,而不是分配必要的 space 来存储结构的内容。
以下是这些函数的正确版本:
struct list *insertnode(struct list *head, char *rom, char opt) //creates a head or creates and adds a new node to the end
{
struct list *new;
if(!head) //if !head, make one
{
new = malloc(sizeof(*new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = NULL;
return new;
}
if(opt == 'o') //if the roman form is optimized, put it in front of all the others
{
new = malloc(sizeof(*new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = head;
return new;
}
head->next = insertnode(head->next, rom, opt); //recursive insertions
return head;
}
struct branch *insertbranch(struct branch *root, int dec, char *rom, char opt) //creates a root or creates and adds a new branch
{
struct branch *new;
if(!root) //if !root, make a root...
{
new = malloc(sizeof(*new));
new->list = insertnode(new->list, rom, opt);
new->dec = dec;
new->count = 1;
new->right=new->left=NULL;
return new;
}
if(dec<root->dec) root->left = insertbranch(root->left, dec, rom, opt); //branch on the left, recursive
else if(dec>root->dec) root->right = insertbranch(root->right, dec, rom, opt); //branch on the right, recursive
else //if there already is such a branch, increase its count
{
root->count += 1;
root->list = insertnode(root->list, rom, opt);
}
return root;
}
我没有仔细查看每一行代码,但看起来大部分是正确的。从我的快速浏览来看,您唯一的错误是分配的内存少于您使用的内存。写入过去分配的内存的结果是不可预测的,并且可以在程序中很晚才体现出来(例如释放内存时)。这就是为什么未定义的行为很难调试的原因:)
我发现了问题。在递归调用 freetree
之后,我尝试以 free(root->left)
和 free(root->right)
的形式再次释放相同的内存。我现在觉得有点傻。
我正在研究我们的教授给我们准备即将到来的考试的一些旧课文,我 运行 解决了这个问题。
我的任务是从结构如下的文本文件中读取信息:
[十进制数],[罗马数(字符串)],[o或u(优化或未优化的罗马数)]
几千行并将该信息存储在二叉搜索树中,使用十进制数作为键。每个 b运行ch 还必须包含该数字出现的次数和遇到的各种罗马版本的列表,优化的版本位于列表的顶部。 然后释放一切。
我的代码(c):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct list //list node
{
char *rom;
struct list *next;
};
struct branch //main tree branch
{
int count, dec;
struct list *list;
struct branch *right, *left;
};
struct list *insertnode(struct list *head, char *rom, char opt) //creates a head or creates and adds a new node to the end
{
struct list *new;
if(!head) //if !head, make one
{
new = malloc(sizeof(new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = NULL;
return new;
}
if(opt == 'o') //if the roman form is optimized, put it in front of all the others
{
new = malloc(sizeof(new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = head;
return new;
}
head->next = insertnode(head->next, rom, opt); //recursive insertions
return head;
}
struct branch *insertbranch(struct branch *root, int dec, char *rom, char opt) //creates a root or creates and adds a new branch
{
struct branch *new;
if(!root) //if !root, make a root...
{
new = malloc(sizeof(new));
new->list = insertnode(new->list, rom, opt);
new->dec = dec;
new->count = 1;
new->right=new->left=NULL;
return new;
}
if(dec<root->dec) root->left = insertbranch(root->left, dec, rom, opt); //branch on the left, recursive
else if(dec>root->dec) root->right = insertbranch(root->right, dec, rom, opt); //branch on the right, recursive
else //if there already is such a branch, increase its count
{
root->count += 1;
root->list = insertnode(root->list, rom, opt);
}
return root;
}
void freelist(struct list *head) //frees list 'head'
{
struct list *tmp;
while(head)
{
tmp = head;
head = head->next;
free(tmp->rom);
free(tmp); // <- OFFENDING LINE
}
return;
}
void freetree(struct branch *root) //frees tree 'root'
{
if(!root) return;
freetree(root->left); //recursive!
freetree(root->right); //and on the right
free(root->right);
free(root->left);
freelist(root->list);
free(root);
return;
}
int main()
{
struct branch *root;
struct list *list;
int dec, i=1, n;
char rom[30], opt;
FILE *file = fopen("rom.csv", "r");
if(!file) //is the file even there?
return 1;
while(fscanf(file, "%d,%[^,],%c\n", &dec, rom, &opt)==3) //go through the file and fill tree
root = insertbranch(root, dec, rom, opt);
freetree(root);
printf("Goodbye!\n");
return 0;
}
调试后我确定了问题:在函数 "freelist" 中,当它到达命令 "free(tmp)" 时程序中止。我不知道可能是什么原因。我什至检查以确保节点头存在。
感谢您的帮助!
您没有在 insertnode()
和 insertbranch()
中分配正确的内存量。
你需要替换这个:
new = malloc(sizeof(new));
有:
new = malloc(sizeof(*new));
这是因为new
是一个指针。使用 malloc(sizeof(new))
,您只分配 space 来存储指针,而不是分配必要的 space 来存储结构的内容。
以下是这些函数的正确版本:
struct list *insertnode(struct list *head, char *rom, char opt) //creates a head or creates and adds a new node to the end
{
struct list *new;
if(!head) //if !head, make one
{
new = malloc(sizeof(*new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = NULL;
return new;
}
if(opt == 'o') //if the roman form is optimized, put it in front of all the others
{
new = malloc(sizeof(*new));
new->rom = malloc(sizeof(char)*(strlen(rom)+1));
strcpy(new->rom, rom);
new->next = head;
return new;
}
head->next = insertnode(head->next, rom, opt); //recursive insertions
return head;
}
struct branch *insertbranch(struct branch *root, int dec, char *rom, char opt) //creates a root or creates and adds a new branch
{
struct branch *new;
if(!root) //if !root, make a root...
{
new = malloc(sizeof(*new));
new->list = insertnode(new->list, rom, opt);
new->dec = dec;
new->count = 1;
new->right=new->left=NULL;
return new;
}
if(dec<root->dec) root->left = insertbranch(root->left, dec, rom, opt); //branch on the left, recursive
else if(dec>root->dec) root->right = insertbranch(root->right, dec, rom, opt); //branch on the right, recursive
else //if there already is such a branch, increase its count
{
root->count += 1;
root->list = insertnode(root->list, rom, opt);
}
return root;
}
我没有仔细查看每一行代码,但看起来大部分是正确的。从我的快速浏览来看,您唯一的错误是分配的内存少于您使用的内存。写入过去分配的内存的结果是不可预测的,并且可以在程序中很晚才体现出来(例如释放内存时)。这就是为什么未定义的行为很难调试的原因:)
我发现了问题。在递归调用 freetree
之后,我尝试以 free(root->left)
和 free(root->right)
的形式再次释放相同的内存。我现在觉得有点傻。