Is there a more efficient way to extract the file name of a file?

Question

The code bellow extracts the file name of a file using its path.

#include <iostream>
#include <vector>

using namespace std;

int main() {
    cout << "Program operating..." << endl;

    string s = "C:\\Users\\user\\Pictures\\Strategic_Offense_Logo_1";
    string name;
    for (unsigned int i = s.size() - 1; i > 0; i--) {
        if (s[i] == '\\') {
            for (unsigned int j = i + 1; j < s.size(); j++) {
                if (s[j] == '.' || j == s.size() - 1) {
                    if (j == s.size() - 1)
                        j++;
                    vector<char> v(j - i - 1);
                    unsigned int l = 0;
                    for (unsigned int k = i + 1; k < j; k++) {
                        v[l++] = s[k];
                    }
                    char* vectorAsArray = &v[0];
                    name = vectorAsArray;
                    name.resize(v.size());
                    break;
                }
            }
            break;
        }
    }
    cout << name << endl;
}

#include <iostream>
#include <vector>

using namespace std;

int main() {
    cout << "Program operating..." << endl;

    string s = "C:\\Users\\user\\Pictures\\Strategic_Offense_Logo_1.png";
    string name;
    for (unsigned int i = s.size() - 1; i > 0; i--) {
        if (s[i] == '\\') {
            for (unsigned int j = i + 1; j < s.size(); j++) {
                if (s[j] == '.' || j == s.size() - 1) {
                    if (j == s.size() - 1)
                        j++;
                    vector<char> v(j - i - 1);
                    unsigned int l = 0;
                    for (unsigned int k = i + 1; k < j; k++) {
                        v[l++] = s[k];
                    }
                    char* vectorAsArray = &v[0];
                    name = vectorAsArray;
                    name.resize(v.size());
                    break;
                }
            }
            break;
        }
    }
    cout << name << endl;
}

Is there a more efficient way to do this?

Purpose: I am making a texture class and I want to refer the texture by its name instead of its path or a made up ID.

Note: library does not work at all with the visual studio compiler. If you know how to fix it or have an alternative solution, please post.

Have you considered using the file system header? It is fully functional in VS2019 and 2022 — ChrisMM, Jan 25 '22 at 22:21
You must set the langauge setting to use C++17 or later via: Project > Properties > C/C++ > Language > C++ Language Standard > C++17, C++20, or Preview C++ latest — Casey, Jan 25 '22 at 22:43

score 6 · Answer 1 · answered Jan 25 '22 at 22:21

6

Use std::filesystem::path::filename.

That said, you should refer to the texture by its fully qualified path or a unique ID; otherwise, you will get collisions for textures with the same name in different folders.

answered Jan 25 '22 at 22:21

Casey

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No. This solution does not work because std::filesystem::path::filename is not even recognized by visual studio. Otherwise I would be using it. Also, the reason why I want to refer it by its name instead of by its path is because the header file for filesystem is not found when I try to include it. I looked into many examples online and everything I found on this header does not work. I am not the only one with this problem. – Christopher Barrios Agosto Jan 25 '22 at 22:25
What version of visual studio are you using? [You may need to instruct VS to use C++17.](https://stackoverflow.com/questions/41308933/how-to-enable-c17-compiling-in-visual-studio) – user4581301 Jan 25 '22 at 22:28
If C++17 really isn't available, [give this a try](https://learn.microsoft.com/en-us/windows/win32/api/shlwapi/nf-shlwapi-pathstrippatha). – user4581301 Jan 25 '22 at 22:29
I am using visual studio 2022. Visual Studio 2019 has the same issue. What are you doing different that filesystem works for you but does not work for me? – Christopher Barrios Agosto Jan 25 '22 at 22:31
You mentioned something that I answered lots of question. The version of C++. Apparently, Visual Studio defaults to C++ version 14. To be able to access all the features of the filesystem library one has to go into the solution properties and change the C++ version. So dumb. – Christopher Barrios Agosto Jan 25 '22 at 22:38
That's it: Project > Properties > Configuration Properties > General > C++ Language Standard > /std::c++17. – rturrado Jan 25 '22 at 23:27

Remy Lebeau · Answer 2 · 2022-01-25T23:43:43.227

The simplest solution is to use the <filesystem> library in C++17 onward, specifically the std::filesystem::path class and its stem() method:

Returns the filename identified by the generic-format path stripped of its extension.

#include <iostream>
#include <filesystem>

namespace fs = std::filesystem;
using namespace std;

int main() {
    cout << "Program operating..." << endl;

    //fs::path p = "C:" / "Users" / "user" / "Pictures" / "Strategic_Offense_Logo_1.png";
    fs::path p = "C:\\Users\\user\\Pictures\\Strategic_Offense_Logo_1.png";
    fs::path name = p.stem();

    cout << name << endl;
}

However, if that is not an option, there is a simple solution using the available methods of std::string, such as rfind() and substr():

#include <iostream>
#include <string>

using namespace std;

int main() {
    cout << "Program operating..." << endl;

    string s = "C:\\Users\\user\\Pictures\\Strategic_Offense_Logo_1.png";

    string name = s.substr(s.rfind('\\') /* or: s.find_last_of("\\/") */ + 1);
    name.resize(name.rfind('.'));

    cout << name << endl;
}

score 1 · Answer 3 · edited Jan 25 '22 at 23:44

1

Visual Studio defaults to C++14. To be able to access <filesystem> and all of its features (like extracting file names), one has to go into the solution properties under C++ and change the version to the latest. Then, and only then, <filesystem> will work, as everyone in the Internet and StackOverflow talks about.

edited Jan 25 '22 at 23:44

Remy Lebeau

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answered Jan 25 '22 at 22:42

Christopher Barrios Agosto

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Pre C++17 `boost::filesystem` provided something _nearly_ compatible with `std::filesystem` before C++17. In fact, it's `std::filesystem` _is_ what boost proposed, but with a few adjustments, got into `std`. – Ted Lyngmo Jan 26 '22 at 01:04

Is there a more efficient way to extract the file name of a file?

3 Answers3