Concatenate strings and numbers

Question

I have a function which takes a const char* argument. I need to concatenate two string literals and an int to pass to this argument. Basically this is what I'm trying to do:

open(const char* filename) {}

void loadFile(int fileID)
{
    open("file" + fileID + ".xml");
}

int main()
{
    loadFile(1);
    return 0;
}

How can I make this work as simply as possible? I tried changing the loadFile function to take a const char* and then doing open(std::string("file").c_str() + fileID + std::string(".xml").c_str()); but then I get error: invalid operands of types 'const char*' and 'const char*' to binary 'operator+' so I'm pretty lost.

Simply as possible you ask? Use [`std::string`](http://en.cppreference.com/w/cpp/string/basic_string). — David G, Nov 17 '13 at 20:40
Use `std::to_string` on the `int` and then `c_str` at the end. — chris, Nov 17 '13 at 20:40
@chris You cannot create a `string` from an `int` without using a stream, formatter or something like that... Direct creation will fail and concatenation will interpret the `int` as a `char`. — Johan, Nov 17 '13 at 20:45
@JeremiahWillcock, Indeed it is, but the OP hasn't disallowed it. — chris, Nov 17 '13 at 20:51
@chris: sorry, I did read `std::string` instead of `std::to_string`. Thanks for the tip, I did not know this one. — Johan, Nov 17 '13 at 20:51

score 4 · Answer 1 · answered Nov 17 '13 at 20:40

4

You need to use something like:

std::ostringstream os;
os << "file" << fileID << ".xml";
open(os.str().c_str());

answered Nov 17 '13 at 20:40

Zan Lynx

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Johan · Answer 2 · 2013-11-17T20:58:15.923

4

You can use the stringstream as stated before or Boost format:

#include <boost/format.hpp>

void loadFile(int fileID)
{
  std::string filename = (boost::format("File%d.xml") % fileID).str();
  open(filename.c_str();
}

edited Nov 17 '13 at 20:58

answered Nov 17 '13 at 20:49

Johan

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Would be more complete with a relevant include and fully qualified access of the `format` function. – Captain Obvlious Nov 17 '13 at 20:59

score 2 · Accepted Answer · answered Nov 17 '13 at 20:52

2

If your compiler supports C++11, you can use std::to_string to get a string representation of a number:

std::string filename = "file" + std::to_string(fileId) + ".xml";

However, if you have Boost available, I think using Boost format, as discussed in Johan's answer, is more readable.

answered Nov 17 '13 at 20:52

Chris W.

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Chosen because I can keep everything within the function call, `open(std::string("file" + std::to_string(fileID) + ".xml").c_str());` which is exactly my goal. – Mike Nov 17 '13 at 20:59

score 0 · Answer 4 · answered Nov 17 '13 at 20:54

0

Use to_string()

open("file" + to_string(fileID) + ".xml");

answered Nov 17 '13 at 20:54

Parag

662
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score 0 · Answer 5 · answered Nov 17 '13 at 22:03

C++ is a superset of c. You could use sprintf:

void loadFile(unsigned int fileID)
{
   const int BUFFER_SIZE = 128;
   char buffer[BUFFER_SIZE];
   sprintf(buffer,"file%u.xml");
   open(buffer);
}

This is portable, should be safe for all passed in (uint) values, etc.

You can also use snprintf(buffer,BUFFER_SIZE,....) if you are concerned about overrunning the buffer.

Concatenate strings and numbers

5 Answers5