Pass a reference to std::ifstream as parameter

Question

I'm trying to write a function with an ifstream& argument.

void word_transform(ifstream & infile)
{
    infile("content.txt");
    //etc
}

which gave me an error:

Type 'ifstream' (aka 'basic_ifstream ') does not provide a call operator.

Can you please me what's wrong?

Functions that take stream arguments usually work with a stream that is already open. For example, `ifstream s("content.txt"); word_transform(s);`. — molbdnilo, Dec 16 '14 at 09:23

jpo38 · Accepted Answer · 2014-12-16T11:13:05.403

call operator is a function like operator()( params ) allowing to use the syntax myObject( params ).

So, when you write infile(...), you are trying to us a call operator.

What you are trying to do is to open a file, use the open method:

void word_transform(ifstream & infile)
{
    infile.open("content.txt",std::ios_base::in);
    if ( infile.is_open() )
        infile << "hello";
    infile.close();
}

But, as commented, it does not really make sense to pass infile reference to such a function. You may consider:

void word_transform(istream& infile)
{
    infile << "hello";
}

int main()
{
    ifstream infile;
    infile.open("content.txt",std::ios_base::in);
    if ( infile.is_open() )
        word_transform( infile );
    infile.close();
    return 0;
}

Or:

void word_transform()
{
    ifstream infile;
    infile.open("content.txt",std::ios_base::in);
    if ( infile.is_open() )
        infile << "hello";
    infile.close();
}

int main()
{
    word_transform();
    return 0;
}

Your last comment is very pertinent: if he is going to open and close the file in the function, the `std::ifstream` should be local to the function; it serves no purpose to have it as an argument. A more frequent idiom, however, would be to have it opened and closed by the caller, and to pass a `std::istream&` argument (rather than an `std::ifstream&`). — James Kanze, Dec 16 '14 at 10:44

score 3 · Answer 2 · answered Dec 16 '14 at 09:23

You attempt to call operator() on your parameter. That will not work. Are you trying to open a file? If you get an ifstream as parameter, it should be open from the start because you opened it outside your function. Passing a stream and then opening it inside your function does not make sense.

void word_transform(std::ifstream& infile)
{
    // read something from infile
}

int main()
{
   std::ifstream file("content.txt");

   // error checks

   word_transform(file);

   return 0;
}

score 1 · Answer 3 · answered Dec 16 '14 at 09:23

1

infile("content.txt");

Note that this would try to call operator() on already created object of type infile. As no such operator exists from ifstream , you got an error.

Rather you should do:-

infile.open("content.txt");

answered Dec 16 '14 at 09:23

ravi

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score 0 · Answer 4 · answered Dec 16 '14 at 09:23

0

To my understanding, the call operator is operator(), which apparently is not defined for ifstream. You have to do something different with the argument of word_transform.

answered Dec 16 '14 at 09:23

Codor

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score 0 · Answer 5 · answered Dec 16 '14 at 09:32

Typically, (references to) streams are passed to functions for I/O. This means the function should take std::istream or std::ostream argument, but not std::ifstream or std::ofstream. This way your function can be used with any stream object derived from std::istream, including std::cin, std::istrstream, and std::ifstream.

As nvoigt said, passing an std::ifstream to a function for opening makes little sense. It is definitely not clear to the caller that its stream may be closed and opened to another file.

Pass a reference to std::ifstream as parameter

5 Answers5

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