我构造了一个用于处理某个文件形式的类,它的构造函数遍历文件并搜索我需要的关键信息-想法是字符写在多行上,我想读取每行的第一个字符,每行的第二个字符,依此类推。
我有下面的构造函数和定义(可能很可怕——这是我第一次用C++写任何严肃的东西),
class AlignmentStream{
private:
const char* FileName;
std::ifstream FileStream;
std::vector<int> NamesStart;
std::vector<int> SequencesStart;
std::vector<int> SequenceLengths;
int CurrentPosition;
int SequenceNum;
public:
AlignmentStream(const char* Filename);
std::vector<int> tellSeqBegins();
std::vector<int> tellNamesStart();
std::vector<int> tellSequenceLengths();
int getSequenceNum();
AlignedPosition get();
};
AlignmentStream::AlignmentStream(const char* Filename)
{
FileName = Filename;
FileStream.open(FileName);
std::cout << "Filestream is open: " << FileStream.is_open() << std::endl;
std::cout << "Profiling the alignment file..." << std::endl;
if (FileStream.is_open() == false)
throw StreamClosed(); // Make sure the stream is indeed open else throw an exception.
if (FileStream.eof())
throw FileEnd();
char currentchar;
// Let's check that the file starts out in the correct fasta format.
currentchar = FileStream.get();
if (FileStream.eof())
throw FileEnd();
if (currentchar != '>')
throw FormatError();
NamesStart.push_back(FileStream.tellg());
bool inName = true;
bool inSeq = false;
int currentLength = 0;
while(!FileStream.eof()){
while (!FileStream.eof() && inName == true) {
if (currentchar == 'n') {
inName = false;
inSeq = true;
SequencesStart.push_back(FileStream.tellg());
} else {
currentchar = FileStream.get();
}
}
while (!FileStream.eof() && inSeq == true) {
if (currentchar == '>') {
inName = true;
inSeq = false;
NamesStart.push_back(FileStream.tellg());
} else {
if (currentchar != 'n') {
currentLength++;
}
currentchar = FileStream.get();
}
}
SequenceLengths.push_back(currentLength); // Sequence lengths is built up here - (answer to comment)
currentLength = 0;
}
SequenceNum = (int)SequencesStart.size();
// Now let's make sure all the sequence lengths are the same.
std::sort(SequenceLengths.begin(), SequenceLengths.end());
//Establish an iterator.
std::vector<int>::iterator it;
//Use unique algorithm to get the unique values.
it = std::unique(SequenceLengths.begin(), SequenceLengths.end());
SequenceLengths.resize(std::distance(SequenceLengths.begin(),it));
if (SequenceLengths.size() > 1) {
throw FormatError();
}
std::cout << "All sequences are of the same length - good!" << std::endl;
CurrentPosition = 1;
FileStream.close();
}
很抱歉它是一个相当大的块,不管怎样,构造函数都会逐个字符地读取每一行的起点。然后,get函数(未显示)遍历并查找每一行的开头+还有多少个字符要到达正确的字符-由成员变量CurrentPos给定。然后,它构造了我的另一个自定义对象AlignedPosition并返回它
AlignedPosition AlignmentStream::get()
{
std::vector<char> bases;
for (std::vector<int>::iterator i = SequencesStart.begin(); i != SequencesStart.end(); i++) {
// cout messages are for debugging purposes.
std::cout << "The current filestream position is " << FileStream.tellg() << std::endl;
std::cout << "The start of the sequence is " << *i << std::endl;
std::cout << "The position is " << CurrentPosition << std::endl;
FileStream.seekg((int)(*i) + (CurrentPosition - 1) );
std::cout << "The Filestream has been moved to " << FileStream.tellg() << std::endl;
bases.push_back(FileStream.get());
}
CurrentPosition++;
//this for loop is just to print the chars read in for debugging purposes.
for (std::vector<char>::iterator i = bases.begin(); i != bases.end(); i++) {
std::cout << *i << std::endl;
}
return AlignedPosition(CurrentPosition, bases);
}
正如您所看到的,第一个循环遍历每一行的起始位置+CurrentPosition,然后获取char并将其推回到向量上,该向量被传递给我的AlignedPosition构造函数,其他一切都是用于调试的消息。然而,在执行时,我看到了这一点:
eduroam-180-37:libHybRIDS wardb$ ./a.out
Filestream is open: 1
Profiling the alignment file...
All sequences are of the same length - good!
SeqNum: 3
Let's try getting an aligned position
The current filestream position is -1
The start of the sequence is 6
The position is 1
The Filestream has been moved to -1
The current filestream position is -1
The start of the sequence is 398521
The position is 1
The Filestream has been moved to -1
The current filestream position is -1
The start of the sequence is 797036
The position is 1
The Filestream has been moved to -1
?
?
?
Error, an invalid character was present
Couldn't get the base, caught a format error!
简而言之,我看到的是文件流的位置是-1,并且在使用查找时不会改变。这会导致在我的AlignedPosition构造函数中引发无效字符和异常。这是否与已经在我的构造函数中导航到文件末尾有关?为什么我在输入流中的位置一直保持在-1?
谢谢,本。
如果您在流上获得文件结尾,seekg
可能无法清除它。您需要首先在流上调用clear()
。由于您一直读到EOF,您可能需要调用clear()
。(参考:en.wikipedia.org/wiki/Seekg)