我有一个结构很简单的文件:
{number}\#
{some_text_with_lines}
\#
问题是我希望程序读取此文件中的每个条目。这意味着此结构不止一次出现。
所以,我的第一个想法是创建一个结构数组,某种struct my_struct abc[MAX];
。 MAX
是预处理器定义。
我希望fscanf
将{number}存储为数组,将{some_text_with_lines}存储为struct数组中的字符串。所以每个数组都有自己的字符串。
这个想法是fscanf读取以下格式:%d#\r\n%s\r\n#r\n
,并使用第一个整数%d 作为数组的编号。像& abc [%d]之类的东西
虽然我知道它的语法错误,但我不知道如何读取这个数字%d并将其用作数组的编号。此外,%s存在问题,它没有\ 0?
所以,我需要一些帮助,先谢谢。
答案 0 :(得分:1)
做这样的事情:
int index;
char array[1024][1024]; // big array :)
while(fscanf(file, "%d#\r\n", &index) == 1) {
fscanf(file, "%s\r\n", array[index]);
int c;
while ((c = fgetc(file)) != EOF && c != '\r') { }
fgetc(file); // read '\n'
}
答案 1 :(得分:1)
此代码有效。它将接受文件中的任意段落列表,并且每个段落中的行集也允许具有任意长度。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct data
{
int number;
char **lines;
int num_lines;
int max_lines;
};
int main(void)
{
struct data *info = 0;
int num_entries = 0;
int max_entries = 0;
char line[4096];
while (fgets(line, sizeof(line), stdin) != 0)
{
int number;
char hash;
if (sscanf(line, "%d%c", &number, &hash) != 2 || hash != '#')
{
fprintf(stderr, "Format error (number# expected): %s", line);
return(1);
}
if (max_entries >= num_entries)
{
int new_size = (max_entries + 2) * 2;
struct data *new_info = realloc(info, new_size * sizeof(*new_info));
if (new_info == 0)
{
fprintf(stderr, "Out of memory\n");
return(1);
}
info = new_info;
max_entries = new_size;
}
struct data *curr = &info[num_entries];
curr->number = number;
curr->lines = 0;
curr->max_lines = 0;
curr->num_lines = 0;
while (fgets(line, sizeof(line), stdin) != 0)
{
char *p = strchr(line, '\n');
if (p == 0)
{
fprintf(stderr, "Format error: no newline? (%s)\n", line);
return(1);
}
*p = '\0';
if (strcmp(line, "#") == 0)
break;
if (curr->max_lines >= curr->num_lines)
{
int new_size = (curr->max_lines + 2) * 2;
char **new_lines = realloc(curr->lines, new_size * sizeof(*new_lines));
if (new_lines == 0)
{
fprintf(stderr, "Out of memory\n");
return(1);
}
curr->lines = new_lines;
curr->max_lines = new_size;
}
curr->lines[curr->num_lines] = strdup(line);
if (curr->lines[curr->num_lines] == 0)
{
fprintf(stderr, "Out of memory\n");
return(1);
}
curr->num_lines++;
}
num_entries++;
}
for (int i = 0; i < num_entries; i++)
{
printf("%d#\n", info[i].number);
for (int j = 0; j < info[i].num_lines; j++)
printf(" %d: %s\n", j, info[i].lines[j]);
}
return(0);
}
给定输入文件:
13#
Unlucky for some
#
20121221#
The end of the world?
No, it seems that we survived yet another doomsday!
#
18#
More lines,
And more lines still.
The verse is weird.
The terse is worse.
#
19#
As for one,
Then another,
It is still too short
For comfort,
But the fifth line shall trigger
an extra reallocation.
#
20#
All for one,
And one for all!
The Three Musketeers?
Nay, D'Artagnan, the Four Musketeers.
Yahoo! Bing? Google?
#
它给出了输出:
13#
0: Unlucky for some
20121221#
0: The end of the world?
1: No, it seems that we survived yet another doomsday!
18#
0: More lines,
1: And more lines still.
2: The verse is weird.
3: The terse is worse.
19#
0: As for one,
1: Then another,
2: It is still too short
3: For comfort,
4: But the fifth line shall trigger
5: an extra reallocation.
20#
0: All for one,
1: And one for all!
2: The Three Musketeers?
3: Nay, D'Artagnan, the Four Musketeers.
4: Yahoo! Bing? Google?
代码中无疑存在改进的空间,但它似乎有效(尽管它在退出之前不会释放内存)。它假定有一个函数char *strdup(const char *str)
为它给出的字符串副本分配足够的空间。