如何计算C中的一些分布缩写 [英] How to calculate some distribution abbreviation in C

问题描述

我有一个包含多个文件的目录。每个文件都有一列值。

我编写一个代码打开目录并逐个打开文件，计算Min，Max和Average然后转到下一个文件。

但有一个问题我无法得到它？!!

这是我的代码：

Hi,
I have a directory which has multiple files. Each file has a column of values.
I write a code which open the directory and open files one by one, calculate Min, Max and Average and then goes to the next file.
But there is a problem I can't get it?!!
Here is my code:

static void scan_dir(const char *dir)
    {
        struct dirent * entry;
        DIR *d = opendir( dir );
    
        if (d == 0) {
            perror("opendir");
            return;
        }
    
        while ((entry = readdir(d)) != 0)
        {
            FILE *sensor;
            int num, min, max, sum, count, first;
            sensor = fopen( entry->d_name, "r");
            if (sensor != NULL)
            {
                for (sum = count = max = min = first = 0;
                fscanf(sensor, "%d", &num) == 1; sum += num, ++count)
                {
                    if (!first)
                    {
                        min = max = num; first = 1;
                    }
                    else if (num > max) max = num;
                    else if (num < min) min = num;
    	    }
               
                printf("count = %d, min = %d, max = %d, avg = %.1lf\n", count, min, max, sum / (double) count);
            }
        }
        closedir(d);
    }
    
    
    int main(int argc, char ** argv)
    {
        scan_dir(argv[1]);
        return 0;
    }

提前致谢，

Kambiz

Thanks in advance,
Kambiz

推荐答案

我们不做你的功课，所以我们不打算为你解决 - 这是一项重要的开发技巧，叫做调试，你只能通过使用而不是通过观察来学习。



所以使用你的调试器。

在scan_dir函数的第一行放一个断点，然后逐步查看代码，看看究竟发生了什么。弄清楚每一行在执行之前应该发生什么，并比较发生了什么。如果它达到了你的预期，那么它很好，继续下一步。如果没有，为什么不呢？出了什么问题？是代码还是您的期望？



尝试一下：这是一项有价值的技能，在你完成复杂的任务之前，最好在这样的简单任务中学习！



但我建议如果你改变你的编码风格会容易得多：将scan_dir分成不同的任务，所以它处理一个目录，并为每个文件调用一个单独的函数。然后每个文件通过另一种方法处理一组行。

这样你就简化了代码并分离了问题，它应该更容易调试。



[edit]

原始OP问题：

我有一个目录有多个文件。

每个文件都有一列值。

我写一个代码打开目录并逐个打开文件，计算最小值，最大值和平均值¬

然后转到下一个文件。

但有一个问题我无法得到它？

！

！

这是我的代码：

We don't do your homework, so we aren't going to fix it for you - that's an important development skill called debugging, which you only learn by using, not by watching.

So use your debugger.
Put a breakpoint on the first line of the scan_dir function, and step through the code looking at what exactly is going on. Work out what should happen as a result of each line before it is executed, and compare what did happen to that. If it did what you expected, then it;s fine, move on to the next. If it didn't, why not? What was the problem? Was it the code, or your expectations?

Try it: this is a valuable skill, and one which is best learned on simple tasks like this before you get to complicated tasks!

But I'd suggest that it would be a lot easier if you change your coding style: break scan_dir into separate tasks, so it handles a directory, and calls a separate function for each file. Then each file processes a set of lines via another method.
That way you are simplifying the code and separating the concerns and it should be easier to debug.

[edit]
Original OP question:
Hi, I have a directory which has multiple files.
Each file has a column of values.
I write a code which open the directory and open files one by one, calculate Min, Max and Average ¬
and then goes to the next file.
But there is a problem I can't get it?
!
!
Here is my code:

static void scan_dir(const char *dir)
{
struct dirent * entry;
DIR *d = opendir( dir );
 
if (d == 0) {
perror("opendir");
return;
}
 
while ((entry = readdir(d)) != 0)
{
printf("%s\n", entry->d_name);
FILE *sensor;
int num, min, max, sum, count, first;
sensor = fopen( entry->d_name, "r");
if (sensor != NULL)
{
for (sum = count = max = min = first = 0;
fscanf(sensor, "%d", &num) == 1; sum += num, ++count)
{
if (!first)
{
min = max = num; first = 1;
}
else if (num > max) max = num;
else if (num < min) min = num;
}
 
printf("count = %d, min = %d, max = %d, avg = %.1lf\n", count, min, max, sum / (double) count);
fclose(sensor);
}
else
{
printf("Unable to read file '%s'\n", entry->d_name );
}
}
closedir(d);
}
 
 
int main(int argc, char ** argv)
{
scan_dir(argv[1]);
return 0;
}

这是输出：

and here is the output:

1 file 1
2 count = 41, min = 1254, max = 1361, avg = 1298.3
3 ..
4 count = 0, min = 0, max = 0, avg = -nan
5 file 2
6 count = 41, min = 1865245, max = 1936860, avg = 1875714.9
7 file 3
8 count = 41, min = 0, max = 0, avg = 0.0
9 .
10 count = 0, min = 0, max = 0, avg = -nan
11 file 4
12 count = 41, min = 1, max = 1, avg = 1.0

以上，我不知道为什么印有3,4,9,10的线？ br $> b $ b！

实际上它们甚至与任何文件无关！



所以有人可以帮我纠正代码？

[/ edit]

问题在于：此脚本读取目录中的所有文件;所以如果有一些隐藏文件，输出就不是我的预期。所以我在我的if命令中放了一个过滤器来读取.txt文件; - ）

The problem was that: This script read all files within the directory; So if there is some hidden files, the output is not what I expected. So I put a filter in my "if" command to just read the ".txt" files ;-)

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