Extract file name from full path in C using MSVS2005

Question

In a C program, I have a file path in a string (specifically, this is the exe name stored in argv[0]). I would like to extract the file name and di

Answer 1

For reference, here's the code I implemented, supposedly Win/Unix compatible:

    char *pfile;
    pfile = argv[0] + strlen(argv[0]);
    for (; pfile > argv[0]; pfile--)
    {
        if ((*pfile == '\\') || (*pfile == '/'))
        {
            pfile++;
            break;
        }
    }

Answer 2

This is the poorman's version:

char *p, *s = args[0]; // or any source pathname
...
p = strchr(s, '\\'); // find the 1st occurence of '\\' or '/'
// if found repeat the process (if not, s already has the string)
while(p) {
  s = ++p; // shift forward s first, right after '\\' or '/'
  p = strchr(p, '\\'); // let p do the search again
}
// s now point to filename.ext
...

note: for _TCHAR, use _tcschr rather that strchr

strchr is similar to: while((*p) && (*p != '\\')) p++; with safebelt NULL return if chr is not found. So if you really don't like to depend on another lib you can use this:

char *p, *s = args[0];
...
p = s; // assign to s to p
while(*p && (*p != '\\')) p++;
while(*p) { // here we have to peek at char value
  s = ++p;
  while (*p && (*p != '\\')) p++;
}
// s now point to filename.ext
...

Any lower than that, use asm instead..

Answer 3

If you wanted a function from libc:

#include <unistd.h>

char * basename (const char *fname);

Answer 4

Not really, just locate the last backslash in the path. Anything after that is the filename. If nothing is after that, the path specifies a directory name.

// Returns filename portion of the given path
// Returns empty string if path is directory
char *GetFileName(const char *path)
{
    char *filename = strrchr(path, '\\');
    if (filename == NULL)
        filename = path;
    else
        filename++;
    return filename;
}

Answer 5

Use _splitpath(). See https://docs.microsoft.com/en-us/cpp/c-runtime-library/reference/splitpath-wsplitpath?redirectedfrom=MSDN&view=vs-2019 for details