Odtnhj

I need to read all files from the folder and write to buffer. All files have the same name except the last portion (file_0000.mdf, file_0001.mdf,....file_9999.mdf). How i can read all the files? testFolder contains all the files. If i go with for_loop,it will start with 0 but my test_files start from 0000 and so on. Also, i need file size of each individual test file. My logic is wrong, but i do not know how to fix it. Some updated code is also given below the first approach.

#include <iostream>
#include <fstream>
#include <string>
int main()

std::string path="C:\testFolder\";
std::string constName="file_";
std::string lastName = ".mdf";
std::fstream InputStream;
std::string fileWithPath;

for (int i=0; i <9999;i++)

 fileWithPath=path+constName+std::to_string(static_cast<long long> 
 (i)+lastName;
 InputStream.open(fileWithPath,std::ios::binary);
 long InputFileSize= InputStream.tellg();
 

 return 0;
 

Short update by using boost/filesystem. Need comments on this approach.

#include <boost/filesystem.hpp>
#include <boost/range/iterator_range.hpp>

std::string SourceFolder="C:\testFolder\";
path mDirectory(SourceFolder);

std::cout<<"Directory includes the following files"
if(is_directory(mDirectory)){
for(auto testFile=mDirectory.begin();testFile!=mDirectory.end();testFile++)
 std::cout<< testFile->string()<<std::endline; 

Plain integers doesn't have leading zeros. To get leading zeros you need to use some other way of formatting your file-names. For example by using std::ostringstream and standard I/O manipulators like std::setw and std::setfill:

std::ostringstream oss;
oss << path << constName << std::setw(4) << std::setfill('0') << i << lastName;
fileWithPath = oss.str();

You may use FindFirstFile() and FindNextFile() functions scanning the files in the directory using wildcard, i.e. "C:Datafile_???.mdf".
Returned WIN32_FIND_DATA will contain also a file size.

Take a look on the complete example "Listing the Files in a Directory".

Once a file is listed, you may read its content with ifstream as usually.