I am trying to figure out how the program /bin/ls -l calculates the total size (block count) for a directory. By this I mean the output total number that it prints right before the directory contents.

There is a similar question here: https://stackoverflow.com/questions/7401704/what-is-that-total-in-the-very-first-line-after-ls-l but it doesn't fully answer the question nor explain exactly how it is calculated.

I've tried adding the numbers of 512B blocks allocated for all the (non-hidden) files in a directory.  Here is how I am trying to go about it (in C):

 int getBlockSize(char* directory) {
   int size = 0;

   DIR *d;
   struct dirent *dir;
   struct stat fileStat;
   d = opendir(directory);
   if (d) {
       while ((dir = readdir(d)) != NULL) {
           if (dir->d_name[0] != '.') { // Ignore hidden files
               // Create the path to stat
               char info_path[PATH_MAX + 1];
               strcpy(info_path, directory);
               if (directory[strlen(directory) - 1] != '/')
                   strcat(info_path, "/");
               strcat(info_path, dir->d_name);

               stat(info_path, &fileStat);

               size += fileStat.st_blocks;

   return size;

However this is giving me a much different number compared to the ls command.

What is 'wrong' with my approach?  How does ls compute the total?


To test I made a folder which contains files test_file1.txt and test_file2.txt each containing the text Hello World!. When I run ls -l I get the following output

total 1
-rw-------. 1 aaa111 ugrad 13 Oct 27 13:17 test_file1.txt
-rw-------. 1 aaa111 ugrad 13 Oct 27 13:17 test_file2.txt

However when I run my code using the method above I get

total 2
-rw-------. 1 aaa111 ugrad 13 Oct 27 13:17 test_file1.txt
-rw-------. 1 aaa111 ugrad 13 Oct 27 13:17 test_file2.txt 

The ls default on Linux is to present the total line in 1K blocks, so your result is correct since you're counting 512-byte blocks.

You can force ls to count in 512-byte blocks using POSIXLY_CORRECT:


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