Creating file names automatically C++

Question

I'm trying to write a program in C++, which creates some files (.txt) and writes down the result in them. The problem is that an amount of these files is not fixed at the beginning and only appears near the end of the program. I would like to name these files as "file_1.txt", "file_2.txt", ..., "file_n.txt", where n is an integer.

I can't use concatenation because the file name requires type "const char*", and I didn't find any way to convert "string" to this type. I haven't found any answer through the Internet and shall be really happy if you help me.

Have you looked at the c_str function? en.cppreference.com/w/cpp/string/basic_string/c_str — R. Martinho Fernandes
– R. Martinho Fernandes, Commented Oct 28, 2012 at 13:00
Just because the filename needs to be a const char* in the end doesn't mean you're stuck with it for your whole program. You should use std::string as much as you can, then obtain a const char* from it only when you really it. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Oct 28, 2012 at 13:09

Maverick Lally · Accepted Answer · 2023-01-10 03:11:34Z

7

You can get a const char* from an std::string by using the c_str member function.

std::string s = ...;
const char* c = s.c_str();

If you don't want to use std::string (maybe you don't want to do memory allocations) then you can use snprintf to create a formatted string:

#include <cstdio>
...
char buffer[16]; // make sure it's big enough
snprintf(buffer, sizeof(buffer), "file_%d.txt", n);

n here is the number in the filename.

edited Jan 10, 2023 at 3:11

Maverick Lally

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answered Oct 28, 2012 at 12:59

Peter Alexander

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2 Comments

Peter Alexander Over a year ago

A note on the memory allocations of std::string: many implementations do the small string optimisation where small strings won't allocate memory, however this isn't guaranteed.

Matthieu M. Over a year ago

@PeterAlexander: Well, MSVC's dirkumware version has SSO while gcc's libstdc++ has not. I think that clang's libc++ has SSO too, however since it's barely ported outside of OS X it's of little interest to most.

Darren Smith · Accepted Answer · 2012-10-28 13:20:12Z

7

 for(int i=0; i!=n; ++i) {
     //create name
     std::string name="file_" + std::to_string(i) + ".txt"; // C++11 for std::to_string 
     //create file
     std::ofstream file(name);
     //if not C++11 then std::ostream file(name.c_str());
     //then do with file
 }

edited Oct 28, 2012 at 13:20

Darren Smith

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answered Oct 28, 2012 at 13:04

111111

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Darren Smith · Accepted Answer · 2012-10-28 13:06:12Z

3

... another way to bulid the filename

#include <sstream>

int n = 3;
std::ostringstream os;
os << "file_" << n << ".txt";

std::string s = os.str();

answered Oct 28, 2012 at 13:06

Darren Smith

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Software_Designer · Accepted Answer · 2012-10-28 19:12:58Z

1

Sample code:

#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
using namespace std;

string IntToStr(int n) 
{
    stringstream result;
    result << n;
    return result.str();
}

int main () 
{
    ofstream outFile;
    int Number_of_files=20;
    string filename;


  for (int i=0;i<Number_of_files;i++)
  {
        filename="file_" + IntToStr(i) +".txt";
        cout<< filename << "  \n";

        outFile.open(filename.c_str());
        outFile << filename<<" : Writing this to a file.\n";
        outFile.close();
  }


  return 0;
}

answered Oct 28, 2012 at 19:12

Software_Designer

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user2138149 · Accepted Answer · 2021-02-04 13:59:05Z

1

I use the following code for this, you may find this useful.

std::ofstream ofile;

for(unsigned int n = 0; ; ++ n)
{
    std::string fname = std::string("log") + std::tostring(n) + std::string(".txt");

    std::ifstream ifile;
    ifile.open(fname.c_str());

    if(ifile.is_open())
    {
    }
    else
    {
        ofile.open(fname.c_str());
        break;
    }

    ifile.close();
}

if(!ofile.is_open())
{
    return -1;
}

ofile.close();

edited Feb 4, 2021 at 13:59

answered Aug 16, 2015 at 14:24

user2138149

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