C++ – How to “normalize” a pathname using boost::filesystem

boostcfilesystems

We are using boost::filesystem in our application. I have a 'full' path that is constructed by concatenating several paths together:

#include <boost/filesystem/operations.hpp>
#include <iostream>
     
namespace bf = boost::filesystem;

int main()
{
    bf::path root("c:\\some\\deep\\application\\folder");
    bf::path subdir("..\\configuration\\instance");
    bf::path cfgfile("..\\instance\\myfile.cfg");

    bf::path final ( root / subdir / cfgfile);

    cout << final.file_string();
}

The final path is printed as:

c:\some\deep\application\folder\..\configuration\instance\..\instance\myfile.cfg

This is a valid path, but when I display it to the user I'd prefer it to be normalized. (Note: I'm not even sure if "normalized" is the correct word for this). Like this:

c:\some\deep\application\configuration\instance\myfile.cfg

Earlier versions of Boost had a normalize() function – but it seems to have been deprecated and removed (without any explanation).

Is there a reason I should not use the BOOST_FILESYSTEM_NO_DEPRECATED macro? Is there an alternative way to do this with the Boost Filesystem library? Or should I write code to directly manipulating the path as a string?

Best Answer

Boost v1.48 and above

You can use boost::filesystem::canonical:

path canonical(const path& p, const path& base = current_path());
path canonical(const path& p, system::error_code& ec);
path canonical(const path& p, const path& base, system::error_code& ec);

http://www.boost.org/doc/libs/1_48_0/libs/filesystem/v3/doc/reference.html#canonical

v1.48 and above also provide the boost::filesystem::read_symlink function for resolving symbolic links.

Boost versions prior to v1.48

As mentioned in other answers, you can't normalise because boost::filesystem can't follow symbolic links. However, you can write a function that normalises "as much as possible" (assuming "." and ".." are treated normally) because boost offers the ability to determine whether or not a file is a symbolic link.

That is to say, if the parent of the ".." is a symbolic link then you have to retain it, otherwise it is probably safe to drop it and it's probably always safe to remove ".".

It's similar to manipulating the actual string, but slightly more elegant.

boost::filesystem::path resolve(
    const boost::filesystem::path& p,
    const boost::filesystem::path& base = boost::filesystem::current_path())
{
    boost::filesystem::path abs_p = boost::filesystem::absolute(p,base);
    boost::filesystem::path result;
    for(boost::filesystem::path::iterator it=abs_p.begin();
        it!=abs_p.end();
        ++it)
    {
        if(*it == "..")
        {
            // /a/b/.. is not necessarily /a if b is a symbolic link
            if(boost::filesystem::is_symlink(result) )
                result /= *it;
            // /a/b/../.. is not /a/b/.. under most circumstances
            // We can end up with ..s in our result because of symbolic links
            else if(result.filename() == "..")
                result /= *it;
            // Otherwise it should be safe to resolve the parent
            else
                result = result.parent_path();
        }
        else if(*it == ".")
        {
            // Ignore
        }
        else
        {
            // Just cat other path entries
            result /= *it;
        }
    }
    return result;
}

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

Python – How to copy a file in Python

shutil has many methods you can use. One of which is:

from shutil import copyfile
copyfile(src, dst)

# 2nd option
copy(src, dst)  # dst can be a folder; use copy2() to preserve timestamp

Copy the contents of the file named src to a file named dst. Both src and dst need to be the entire filename of the files, including path.
The destination location must be writable; otherwise, an IOError exception will be raised.
If dst already exists, it will be replaced.
Special files such as character or block devices and pipes cannot be copied with this function.
With copy, src and dst are path names given as strs.

Another shutil method to look at is shutil.copy2(). It's similar but preserves more metadata (e.g. time stamps).

If you use os.path operations, use copy rather than copyfile. copyfile will only accept strings.