C++ int to string, concatenate strings [duplicate]

Answer 1

You could use a stringstream to construct the filename.

std::ostringstream filename;
filename << "File - " << n << ".txt";
resultsFile.open(filename.str().c_str());

Answer 2

You can use std::stringstream

std::stringstream ss;
ss << "File - " << n << ".txt";

Since the constructor requires a char pointer, you need to convert it into a char pointer using

ofstream resultsFile(ss.str().c_str());

Answer 3

for itoa function

include <stdlib.h>

consider this link

http://www.cplusplus.com/reference/cstdlib/itoa/

Answer 4

Using std::ostringstream:

std::ostringstream os;
os << "File - "  << nStr << ".txt";
std::ofstream resultsFile(os.str().c_str());

Using std::to_string (C++11):

std::string filename = "File - " + std::to_string(nStr) + ".txt";
std::ofstream resultsFile(filename.c_str());

Answer 5

For itoa, you are likely missing #include <stdlib.h>. Note that itoa is non-standard: the standard ways to format an integer as string as sprintf and std::ostringstream.

ofstream.open() takes a const char*, not std::string. Use .c_str() method to obtain the former from the latter.

Putting it together, you are looking for something like this:

ostringstream nameStream;
nameStream << "File - " << n << ".txt";
ofstream resultsFile(nameStream.str().c_str());

Answer 6

You want to use boost::lexical_cast. You also need to include any needed headers:

#include <boost/lexical_cast>
#include <string>
std::string nStr = boost::lexical_cast<std::string>(n);

then it's simply:

std::string file_name = "File-" + nStr + ".txt";

because std::strng plays nicely with string literals (e.g. ".txt").