c++17 `filesystem` is not a namespace-name

Question

I am wondering why in the following code, the namespace filesystem is not found:

g++ -std=c++17 main.cpp -lstdc++

// #include <filesystem>   <- error, so changed to the following:
#include <experimental/filesystem>

namespace fs = std::filesystem;

int main()
{
    return 0;
}

error:

main.cpp:3:21: error: ‘filesystem’ is not a namespace-name
 namespace fs = std::filesystem;
                     ^
main.cpp:3:31: error: expected namespace-name before ‘;’ token
 namespace fs = std::filesystem;

gcc version 5.4.0 20160609 (Ubuntu 5.4.0-6ubuntu1~16.04.5)

If your compiler is older than the standard you are trying to use, there is only so much compliance you can expect. — Baum mit Augen, Jan 18 '18 at 01:29
Also, to be able to link, you have to add the `-lstdc++fs` option to g++. See notes in https://en.cppreference.com/w/cpp/filesystem — Pierre, Jul 20 '20 at 09:08
Works fine for me in `g++ (Ubuntu 9.3.0-17ubuntu1~20.04) 9.3.0` — Dmitry Messerman, Oct 04 '21 at 15:08

score 21 · Accepted Answer · answered Jan 18 '18 at 01:31

21

GCC 5.4.0 was released in June of 2016; over a year before the C++17 standard was adopted. It and its version of libstdc++ have very limited C++17 support. You can see when GCC added C++17 language features here and when libstdc++ added C++17 standard library features here.

At the time of GCC 5.4's release, the filesystem library was not yet implemented in the std::filesystem namespace. It, along with any other <experimental/...> headers that are included in that version, are in the std::experimental namespace.

answered Jan 18 '18 at 01:31

Miles Budnek

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1

Do you know any solution for that? – Wafula Samuel Nov 20 '18 at 08:58
3

@SamZiggler Any solution for what? If you must use an old compiler, then you'll have to include the appropriate experimental header and use `std::experimental::filesystem`. You could use some preprocessor checks to set things up if you want to be compatible with a wide range of compiler versions. – Miles Budnek Nov 20 '18 at 16:36

score 18 · Answer 2 · edited Sep 09 '20 at 18:31

18

<experimental/..> means experimental namespace:

namespace fs = std::experimental::filesystem;

See: http://en.cppreference.com/w/cpp/experimental/fs/path

edited Sep 09 '20 at 18:31

Olivia Stork

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answered Jan 18 '18 at 01:23

PiotrNycz

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c++17 `filesystem` is not a namespace-name

2 Answers2

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