I am able to create the files such as
f.open("file")
f.open("./path/file")
f.open("../path/file")
but not
f.open("~/path...)
f.open("/path...)
How do I get the absolute paths to work?
By default, std::fstream::open(filename)
opens filename for both input and output. Hence that file must exist and you must
have write permission to it.
In your cases:
f.open("file")
f.open("./path/file")
f.open("../path/file")
you were lucky.
In your case:
f.open("~/path...")
you used the path-element ~, which means $HOME in the shell but just
means ~ in C++.
In the case:
f.open("/path...")
you were unlucky: either the file didn't exist or you didn't have write permission.
If you want to open a file simply for input then either:
use std::ifstream
use std::fstream f; f.open(filename,std::ios_base::in);
If you want to open a file simply for output then either:
use std::ofstream
use std::fstream f; f.open(filename,std::ios_base::out);
With f.open("~/path/....") it is necessary for you to expand the ~ in code. This is typically done by calling getenv("HOME") to retrieve the home directory, and replacing every occurence of '~' in the path with the home directory.
When working with absolute paths, like "/path/whatever", you need to check that f.open() succeeds. There are various reasons that f.open() might fail, including access control (protections that prevent opening a file), the file already being opened by another process, the directory does not exist, etc.
Notably, f.open(), when attempting to open a file within a directory, requires that all directories in the path already exist. It won't exercise some magic to make them exist. If you want that, you need to code it. Bear in mind that each phase (creating directories, etc) might fail.
Related
I am confused by a piece of Python code:
with open('/dev/null', 'w+') as null:
It may be because I do not have knowledge of other Operating Systems, but I thought file names are forbidden to have '/' character. If so, I do not understand how this is a valid command.
Now I do understand that when using the open function in Python, if the file exist in a directory other than the current working directory, one has to prepend the path to the file name argument. However, this does not seem to be the case here because the file name argument for the open function is simply '/dev/null/'. Is 'null' the file name.
Is this related to this:
https://en.wikipedia.org/wiki/Null_device
"in some operating systems, the null device is a device file that discards all data written to it but reports that the write operation succeeded. This device is called /dev/null on Unix and Unix-like systems"
On Unix systems, file name cannot contain forward slash as it's used as directory separator. A file can't have a name of exactly one or two dots as they're used for "current directory" and "parent directory", too. A path starting with a forward slash is an absolute path, going all the way down the directory tree from the root path.
In that code, it opens /dev/null, a special character device that discards everything written to it and reports write success. It's possible that in some cases one wants to discard the output from a specific function, like subprocess.run. In this case, opening a handle to the null device is useful.
I have a medium size program I'm developing using eclipse, I can't show the content because it would be too large. At some point I have some lines of fopen, I'm basically trying to create a new file and printing inside such file a content that has been derived from some processing.
So my line is something like a classic
FILE* f = fopen(filename,"w");
where filename is a char array large enough. However the FILE* returned is 0, and when I check the errno it is number 2, which means that the file doesn't exists.
However that's the point, I was trying to create a new file.
Is there something you could suggest that can I further check?
It's worth to note I'm running this stuff on unix, not windows.
Is the value of filename an absolute path or a relative one? If it's the latter, then probably the process you run hasn't the cwd (current working directory) you think it has.
Try using an absolute path, if that works then
use the chdir system call at the begin of main to set the cwd
Your filename might be incorrectly built, or you have no access to the path, or parts of the path don't exist.
Try a simple and clearly legal path first, like C:\\Temp\\x.x or something similar. if this works, it is not your code, but the filename.
I am polling a directory constantly for files and every time I see a file that meets some certain criteria for reading, I open the file and parse it.
string newLine;
ifstream fileReader;
fileReader.open(filename.c_str());
while(!fileReader.eof())
{
getline(fileReader, newLine);
// do some stuff with the line...
}
filereader.close();
The above code is in a loop that runs every 1 second checking a directory for new files. My issue is that as I am transferring files into the folder for processing, my loop finds the file and passes the name of the file to ifstream who then opens it and tries to parse an incomplete file. How do I make ifstream wait until the file is done being written before it tries to parse the file?
EDIT:
Wanted to better word the issue here since a replier seems to have misunderstood my issue. I have 2 directories
mydirectory/
mydirectoryParsed/
THe way my code works is that my program checks for files in mydirectory/ and when it finds them, parses them and uses the information in the files. No writing to the files are done. Once I am done parsing the file, the file is moved to mydirectoryparsed/
The issue is that when I transfer files over the network into mydirectory/ the ifstream sees these files midtransfer and starts reading them before they finish writing to the directory. How do I make ifstream wait until the file is completely written before parsing it?
Don't transfer the files directly into the directory that your program is watching; instead, transfer them into a different directory on the same drive, and then when the transfer is done, move them into the watched directory. That way, the complete file appears in the watched directory in a single atomic operation.
Alternatively, you could use a naming convention in the watched directory — append a suffix like ".partial" to files that are being transferred, and then rename the file to remove the suffix when the transfer is done. Have your program ignore files whose names end with the suffix.
You're not supposed to open the file every time you write in it. Open it once!
Some pseudo-code for this would be :
1- Open file
2- Get the data you want to write, treat that data
3- Call the write to file function
4- Loop until you have nothing left to write
5- Close de file
I have this code that writes successfully a file:
ofstream outfile (path);
outfile.write(buffer,size);
outfile.flush();
outfile.close();
buffer and size are ok in the rest of code.
How is possible put the file in a specific path?
Specify the full path in the constructor of the stream, this can be an absolute path or a relative path. (relative to where the program is run from)
The streams destructor closes the file for you at the end of the function where the object was created(since ofstream is a class).
Explicit closes are a good practice when you want to reuse the same file descriptor for another file. If this is not needed, you can let the destructor do it's job.
#include <fstream>
#include <string>
int main()
{
const char *path="/home/user/file.txt";
std::ofstream file(path); //open in constructor
std::string data("data to write to file");
file << data;
}//file destructor
Note you can use std::string in the file constructor in C++11 and is preferred to a const char* in most cases.
Rationale for posting another answer
I'm posting because none of the other answers cover the problem space.
The answer to your question depends on how you get the path. If you are building the path entirely within your application then see the answer from #James Kanze. However, if you are reading the path or components of the path from the environment in which your program is running (e.g. environment variable, command-line, config files etc..) then the solution is different. In order to understand why, we need to define what a path is.
Quick overview of paths
On the operating systems (that I am aware of), a path is a string which conforms to a mini-language specified by the operating-system and file-system (system for short). Paths can be supplied to IO functions on a given system in order to access some resource. For example here are some paths that you might encounter on Windows:
\file.txt
\\bob\admin$\file.txt
C:..\file.txt
\\?\C:\file.txt
.././file.txt
\\.\PhysicalDisk1\bob.txt
\\;WebDavRedirector\bob.com\xyz
C:\PROGRA~1\bob.txt
.\A:B
Solving the problem via path manipulation
Imagine the following scenario: your program supports a command line argument, --output-path=<path>, which allows users to supply a path into which your program should create output files. A solution for creating files in the specified directory would be:
Parse the user specified path based on the mini-language for the system you are operating in.
Build a new path in the mini-language which specifies the correct location to write the file using the filename and the information you parsed in step 1.
Open the file using the path generated in step 2.
An example of doing this:
On Linux, say the user has specified --output-path=/dir1/dir2
Parse this mini-language:
/dir1/dir2
--> "/" root
--> "dir1" directory under root
--> "/" path seperator
--> "dir2" directory under dir1
Then when we want to output a file in the specified directory we build a new path. For example, if we want to output a file called bob.txt, we can build the following path:
/dir1/dir2/bob.txt
--> "/" root
--> "dir1" directory under root
--> "/" path separator
--> "dir2" directory under dir1
--> "/" path seperator
--> "bob.txt" file in directory dir2
We can then use this new path to create the file.
In general it is impossible to implement this solution fully. Even if you could write code that could successfully decode all path mini-languages in existence and correctly represent the information about each system so that a new path could be built correctly - in the future your program may be built or run on new systems which have new path mini-languages that your program cannot handle. Therefore, we need to use a careful strategy for managing paths.
Path handling strategies
1. Avoid path manipulation entirely
Do not attempt to manipulate paths that are input to your program. You should pass these strings directly to api functions that can handle them correctly. This means that you need to use OS specific api's directly avoiding the C++ file IO abstractions (or you need to be absolutely sure how these abstractions are implemented on each OS). Make sure to design the interface to your program carefully to avoid a situation where you might be forced into manipulating paths. Try to implement the algorithms for your program to similarly avoid the need to manipulate paths. Document the api functions that your program uses on each OS to the user - this is because OS api functions themselves become deprecated over time so in future your program might not be compatible with all possible paths even if you are careful to avoid path manipulation.
2. Document the functions your program uses to manipulate paths
Document to the user exactly how paths will be manipulated. Then make it clear that it is the users responsibility to specify paths that will work correctly with the documented program behavior.
3. Only support a restricted set of paths
Restrict the path mini-languages your program will accept until you are confident that you can correctly manipulate the subset of paths that meet this set of restrictions. Document this to the user. Error if paths are input that do not conform.
4. Ignore the issues
Do some basic path manipulation without worrying too much. Accept that your program will exhibit undefined behavior for some paths that are input. You could document to the user that the program may or may not work when they input paths to it, and that it is the users responsibly to ensure that the program has handled the input paths correctly. However, you could also not document anything. Users will commonly expect that your program will not handle some paths correctly (many don't) and therefore will cope well even without documentation.
Closing thoughts
It is important to decide on an effective strategy for working with paths early on in the life-cycle of your program. If you have to change how paths are handled later it may be difficult to avoid a change in behaviour that might break the your program for existing users.
Try this:
ofstream outfile;
string createFile = "";
string path="/FULL_PATH";
createFile = path.as<string>() + "/" + "SAMPLE_FILENAME" + ".txt";
outfile.open(createFile.c_str());
outfile.close();
//It works like a charm.
That needs to be done when you open the file, see std::ofstream constructor or open() member.
It's not too clear what you're asking; if I understand correctly, you're
given a filename, and you want to create the file in a specific
directory. If that's the case, all that's necessary is to specify the
complet path to the constructor of ofstream. You can use string
concatenation to build up this path, but I'd strongly recommend
boost::filesystem::path. It has all of the functions to do this
portably, and a lot more; otherwise, you'll not be portable (without a
lot of effort), and even simple operations on the filename will require
considerable thought.
I was stuck on this for a while and have since figured it out. The path is based off where your executable is and varies a little. For this example assume you do a ls while in your executable directory and see:
myprogram.out Saves
Where Saves is a folder and myprogram.out is the program you are running.
In your code, if you are converting chars to a c_str() in a manner like this:
string file;
getline(cin, file, '\n');
ifstream thefile;
thefile.open( ("Saves/" + file + ".txt").c_str() );
and the user types in savefile, it would be
"Saves/savefile.txt"
which will work to get to to get to savefile.txt in your Saves folder. Notice there is no pre-slashes and you just start with the folder name.
However if you are using a string literal like
ifstream thefile;
thefile.open("./Saves/savefile.txt");
it would be like this to get to the same folder:
"./Saves/savefile.txt"
Notice you start with a ./ in front of the foldername.
If you are using linux, try execl(), with the command mv.
Do I always have to specify absolute path for objects instantiated from std::fstream class? In other words, is there a way to specify just relative path to them such as project path?
You can use relative paths as well. But they are relative to the environment you call your executable from.
This is OS dependent but all the major systems behave more or less the same AFAIK.
Windows example:
// File structure:
c:\folder\myprogram.exe
c:\myfile.txt
// Calling command from folder
c:\folder > myprogram.exe
In the above example you could access myfile.txt with "c:/myfile.txt" or "../myfile.txt". If myprogram.exe was called from the root c:\ only the absolute path would work, but instead "myfile.txt" would work.
As Rob Kennedy said in the comments there's really nothing special about paths regarding fstream. But here is a code example using a relative path:
#include <fstream>
int main() {
std::ifstream ifs("../myfile.txt");
... // Do something sensible with the file
}
If you have an .exe file running from C:\Users\Me
and you want to write a file to C:\Users\Me\You\text.txt,
then all what you need to do is to add the current path operator ., so:
std::ifstream ifs(".\\you\\myfile.txt");
will work
You can use relative paths. They're treated the same as relative paths for any other file operations, like fopen; there's nothing special about fstream in that regard.
Exactly how they're treated is implementation-defined; they'll usually be interpretted relative to your process's current working directory, which is not necessarily the same as the directory your program's executable file lives in. Some operating systems might also provide a single working directory shared by all threads, so you might get unexpected results if a thread changes the working directory at the same time another thread tries to use a relative path.
Say you have a src folder directly under your project directory and the src folder contains another tmp_folder folder which contains a txt file named readMe.txt. So the txt file can be read in this way
std::ifstream fin("../src/tmp_folder/readMe.txt");
The behaviour is OS specific. Therefore, the best way to handle this IMHO is to make it somebody else's problem. Read the path to the file to open as a string from the user (e.g: command line argument, config file, env variable etc..) then pass that string directly to the constructor of fstream. Document that this is how your program behaves.
I wrote more about path manipulation here: https://stackoverflow.com/a/40980510/2345997
You can specify a path relative to current directory. On Windows you may call GetCurrentDirectory to retrieve current directory or call SetCurrentDirectory to set current directory. There are also some CRT functions available.
On linux also:
// main.cpp
int main() {
ifstream myFile("../Folder/readme.txt");
// ...
}
Assuming the folder structure is something like this:
/usr/Douments/dev/MyProject/main.cpp
/usr/Documents/dev/MyProject/Folder/readme.txt
What I ended up using was a relative path as identified on How to open a file with relative path in C++? which ended up being:
myFile.open("../Release/frequency.dat", ios::in);
*changing myFile to whatever your variable is.