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? http://en.cppreference.com/w/cpp/string/basic_string/c_str — R. Martinho Fernandes, Oct 28 '12 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, Oct 28 '12 at 13:09

score 7 · Accepted Answer · edited Jan 10 '23 at 03:11

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 '23 at 03:11

Maverick Lally

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2

answered Oct 28 '12 at 12:59

Peter Alexander

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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. – Peter Alexander Oct 28 '12 at 13:07
@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. – Matthieu M. Oct 28 '12 at 14:34

score 7 · Answer 2 · edited Oct 28 '12 at 13:20

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 '12 at 13:20

Darren Smith

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answered Oct 28 '12 at 13:04

111111

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score 3 · Answer 3 · answered Oct 28 '12 at 13:06

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 '12 at 13:06

Darren Smith

2,261
16
16

score 1 · Answer 4 · answered Oct 28 '12 at 19:12

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;
}

FreelanceConsultant · Answer 5 · 2021-02-04T13:59:05.733

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();

Creating file names automatically C++

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