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Home/ Questions/Q 6028989
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T04:49:13+00:00 2026-05-23T04:49:13+00:00

I would like to write a function which takes an fstream , processes it,

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I would like to write a function which takes an fstream, processes it, and fills information in a struct I supply in the second argument.

My problem is that I am confused how to use pointers and fstreams as I get debug errors:

Access violation writing location
0xcccccccc.

Here is the main function:

int main()
{
    keyframe_struct kfstruct;

    string ifile = "filename";
    ifstream fin( ifile, ios::binary );

    load_from_keyframe_file( fin, kfstruct );
    fin.close();

    cout << kfstruct.num_keyframes << endl;

    return 0;
}

And here is the function I try to use for parsing the binary file and filling in the information in the struct kfstruct:

struct keyframe_struct
{
    int num_views;
    int num_keyframes;
    vector<keyframe> keyframes;
};

int load_from_keyframe_file( ifstream &fin, keyframe_struct &kfstruct )
{

    char keyword[100];

    while ( !fin.eof() )
    {
        fin.getline( keyword, 100, 0 );

        if ( strcmp( keyword, "views" ) == 0 )
        {
            fin.read(( char* ) kfstruct.num_views, sizeof( int ) );
        }

        else if ( strcmp( keyword, "keyframes" ) == 0 )
        {
            fin.read(( char* ) kfstruct.num_keyframes, sizeof( int ) );
        }

    }

}

Can you tell me what am I doing wrong? I’m sure I am making some huge errors here as I am just a beginner and I still don’t understand clearly what should I and what should I not do with pointers.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T04:49:14+00:00Added an answer on May 23, 2026 at 4:49 am

    You forgot to take the address of your fields:

    fin.read(( char* ) &kfstruct.num_views, sizeof( int ) );
                      ^^^
    

    [As an aside, note that it’s better from a maintenance point of view to do sizeof(kfstruct.num_views). So if the type ever changes, your code will still work.]

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