创建具有打开文件流的类

Question

我创建了一个应该读取 DNA 序列的类：它包含一个 if 流私有成员：

界面：

class Sequence_stream {
    const char* FileName;
    std::ifstream FileStream;
    std::string FileFormat;

public:
    Sequence_stream(const char* Filename, std::string Format);
    NucleotideSequence get();
};

实施：

Sequence_stream::Sequence_stream(const char* Filename, std::string Format)
{
    FileName = Filename;
    FileStream.open(FileName);
    FileFormat = Format;
    std::cout << "Filestream is open: " << FileStream.is_open() << std::endl;
}

NucleotideSequence Sequence_stream::get()
{
    if (FileStream.is_open())
    {
        char currentchar;
        int basepos = 0;
        std::string name;
        std::vector<Nucleotide> sequence;
        currentchar = FileStream.get();
        if (currentchar == '>' && false == FileStream.eof()) {  // Check that the start of the first line is the fasta head character.
            currentchar = FileStream.get(); // Proceed to get the full name of the sequence. Get characters until the newline character.
            while(currentchar != '\n' && false == FileStream.eof())
            {
                if (true == FileStream.eof()) {
                    std::cout << "The file ends before we have even finished reading in the name. Returning an empty NucleotideSequence" << std::endl;
                    return NucleotideSequence();
                }
                name.append(1, currentchar);
                currentchar = FileStream.get();
            } // done getting names, now let's get the sequence.
            currentchar = FileStream.get();
            while(currentchar != '>' && false == FileStream.eof())
            {
                if(currentchar != '\n'){
                    basepos++;
                    sequence.push_back(Nucleotide(currentchar, basepos));
                }
                currentchar = FileStream.get();
            }
            if(currentchar == '>')
            {
                FileStream.unget();
            }
            return NucleotideSequence(name, sequence);
        } else {
            std::cout << "The first line of the file was not a fasta format description line beginning with '>'. Are you sure the file is of FASTA format?" << std::endl;
            return NucleotideSequence();
        }

    } else {
        std::cout << "The filestream is not open..." << std::endl;
        return NucleotideSequence();
    }
}

但是，如果我测试它：

int main()
{
    std::cout << "Let's try and read in a sequence!" << std::endl;
    std::cout << "First we'll create a stream!" << std::endl;
    Sequence_stream MyDNAStream("~/Dropbox/1_20dd5.fasta", "fasta");
    std::cout << "Done!" << std::endl;
    std::cout << "Now let's try and get a sequence!" << endl;
    NucleotideSequence firstsequence = MyDNAStream.get();
    return 0;
} 

我看到 if 流没有打开：

Let's try and read in a sequence!
First we'll create a stream!
Filestream is open: 0
Done!
The filestream is not open...
logout

[Process completed]

虽然我认为构造函数会打开 if 流。我需要做些什么来纠正这个问题，以便创建对象并包含一个开放流？ （我知道我还没有包含一个析构函数，它将在对象销毁时关闭流）。

谢谢， 本。

Answer 1

您的示例显示 is_open 返回 false。我认为你应该在你的构造函数中检查文件确实是打开的，如果没有，就抛出。

在您的情况下，我怀疑这是由于将“~/Dropbox/1_20dd5.fasta”作为输入参数传递。您是否使用完整路径名进行了测试，没有 ~？我不知道处理真实路径扩展的 C++ 库（如 python 的 os.path）。