我需要打开和读取多个文本文件,这些文件将被存储类似于二叉搜索树结构 注意:在用户输入输入之前,都需要打开它们并存储它们的内容。
我确实需要一些建议,因为我不再看到我的错误了。
然而,我真正没有得到的是如何使其互动? 当我按下A'时,它需要从左侧节点和对面读取文本。
输出结构:
------------------ \ n
\ n
标题\ n
file1 //文件存在 - ' A'
file2 //文件为空/非空 - ' B' \ n
文字\ n
\ n
您的选择(A / B):\ n
这是程序输出应该在开头看的方式: terminal
struct的元素
typedef struct _Elements_
{
char* title_;
struct _Elements_* left;
struct _Elements_* right;
char* text_;
} Elements;
//存在正向初始化
int main (int argc, char* argv[])
{
char returned_value;
Elements element;
if (argc != 2)
{
printf("Usage: ./ass2 [file-name]\n");
return 1;
}
returned_value = openFile(&element, argv[1]);
while(returned_value != ('A' || 'B'))
{
repeatEntry(returned_value); // Function to scan the value
}
// Tricky part! Read from left ???
if(returned_value == 'A')
{
printf("%s",element.left->text_);
}
else
{
printf("%s",element.right->text_);
}
return 0;
}
//这里我想初始化节点元素
//对于每个malloc,我必须确保它成功
Elements* newNode(Elements* element)
{
Elements* newNode = (Elements*)malloc(sizeof(Elements));
if(newNode == NULL)
{
printf("[ERR] Out of memory.\n");
return (void*)2; // void* - to avoid warning of different type?
}
newNode->title_ = NULL;
newNode->left = NULL;
newNode->right = NULL;
newNode->text_ = NULL;
return newNode;
}
//打开文件的功能
char openFile(Elements* element, char* input)
{
// Create new memory if tree is empty
if(element == NULL)
{
return newNode(element);
}
FILE* file_open = fopen(input, "r");
if (file_open == NULL)
{
printf("[ERR] Could not read file %s.\n", input);
return 3;
}
// Local variables to handle the parsing file ?
char line[80];
int lenght_of_the_line; // Was just for me
int line_number = 0; // this also
// Do i really need local variables beside struct ?
char* title_local = NULL;
char* first_file = NULL;
char* second_file = NULL;
char* text_local = NULL;
while(fgets(line, 80, file_open))
{
lenght_of_the_line = strlen(line);
//printf("%d ", lenght_of_the_line);
//printf("%s ", line);
// Get title
if((line[lenght_of_the_line - 1] == '\n') && (line_number == 0))
{
title_local = (char*)malloc(lenght_of_the_line * (sizeof(char)) + 1);
if(title_local == NULL)
{
printf("[ERR] Out of memory.\n");
return 2;
}
strcpy(title_local, line);
printf("%s ", element->title_ = title_local);
printf("\n");
}
// Get 1st file name
else if( (line[lenght_of_the_line - 1] == '\n') && (line_number == 1) )
{
if( (line[lenght_of_the_line - 5] == '.') &&
(line[lenght_of_the_line - 4] == 't') &&
(line[lenght_of_the_line - 3] == 'x') &&
(line[lenght_of_the_line - 2] == 't') )
{
// Allocate enough memory for first file name
first_file = (char*)malloc(lenght_of_the_line * (sizeof(char)) + 1);
if(first_file == NULL)
{
printf("[ERR] Out of memory.\n");
return 2;
}
strcpy(first_file, line);
//name of file to open and store to left node
//Seems to work, since I got no error
openFile(element->left, first_file);
}
}
// Get 2nd file name
else if( (line[lenght_of_the_line - 1] == '\n') && (line_number == 2) )
{
if( (line[lenght_of_the_line - 5] == '.') &&
(line[lenght_of_the_line - 4] == 't') &&
(line[lenght_of_the_line - 3] == 'x') &&
(line[lenght_of_the_line - 2] == 't') )
{
second_file = (char*)malloc(lenght_of_the_line * (sizeof(char)) + 1);
if(second_file == NULL)
{
printf("[ERR] Out of memory.\n");
return 2;
}
strcpy(second_file, line);
openFile(element->right, second_file);
}
}
else
{
text_local = (char*)malloc(lenght_of_the_line * (sizeof(char)) + 1);
if(text_local == NULL)
{
printf("[ERR] Out of memory.\n");
return 2;
}
strcpy(text_local, line);
element->text_ = text_local;
printf("%s", element->text_);
}
// Increase line number for 1
line_number++;
}
//free(line);
fclose(file_open);
printf("\n");
printf("Your choice (A/B)? ");
char user_input;
scanf("%c", &user_input);
// DONT FORGET TO FREE THE MEMORY !
return user_input;
}
答案 0 :(得分:0)
这个
while(returned_value != ('A' || 'B'))
不会做你所期望的。它将返回值与编译器用于表示true
的任何值进行比较
因此,当等待returned_value与true
相同时,您的循环可能看起来是无限循环。
你需要
while((returned_value != 'A') && (returned_value != 'B'))
当returned_value
与'A'或'B'相同时,这将离开循环。
这样就解决了无限循环问题,这可能是阻止文件BST交互的核心问题。
顺便说一下,为了方便用户,我建议也接受'a'或'b'。
为避免在每次输入后“再试一次”,请使用
scanf(" %c", &user_input);
在开头有一个空格,用于忽略前导空格,包括每个输入'A'或'B'后的换行/返回。