How to concat an int to a wchar_t* in C++?

Question

I have to create and write on N files, everyone must have an integer ending to identificate it.

This is my piece of code:

for(int i=0; i<MAX; i++)
{
    uscita.open("nameFile"+i+".txt", ios::out); 
    uscita <<  getData() << endl;
    uscita.close();     
}

And that's what I would like to find in my directory after execution:

nameFile0.txt
nameFile1.txt
nameFile2.txt
...
nameFileMAX.txt

The problem of the above code is that I get the compilin' error:

error C2110: '+' Impossible to add two pointers

If I try to create a string for the name, another problem comes in:

string s ="nameFile"+i+".txt";
uscita.open(s, ios::out);

And the problem is:

error C2664: you cannot convert from string to const wchar_t*

What can I do? How can I create files with different names concating int to wchar_t*?

score 3 · Answer 1 · answered Dec 24 '11 at 16:03

3

You can use std::to_wstring:

#include <string>

// ...

std::wstring s = std::wstring("file_") + std::to_wstring(i) + std::wstring(".dat");

(Then use s.c_str() if you need a C-style wchar_t*.)

answered Dec 24 '11 at 16:03

Kerrek SB

Joe · Accepted Answer · 2011-12-22T21:36:47.443

2

You can use a wstringstream

std::wstringstream wss;
wss << "nameFile" << i << ".txt";
uscita.open(wss.str().c_str(), ios::out);

edited Dec 22 '11 at 21:36

answered Dec 22 '11 at 21:23

Joe

A `wstringstream` would be required here, actually. – Etienne de Martel Dec 22 '11 at 21:24
Nope, it does not work. The compiler says: error C2664, impossible to convert from 'std::basic_string<_Elem,_Traits,_Ax>' to 'const wchar_t *' – DavideChicco.it Dec 22 '11 at 21:28
2

You need to use `.c_str()`, updating answer. – Joe Dec 22 '11 at 21:36

score 0 · Answer 3 · answered Dec 23 '11 at 08:43

0

That is easier and faster:

wchar_t fn[16];
wsprintf(fn, L"nameFile%d.txt", i);
uscita.open(fn, ios::out);

answered Dec 23 '11 at 08:43

vpp

Just a warning. `sprintf`, `wsprintf` and friends often lead to buffer overflows if you're not _really_ careful. (Either now or when the code is being maintained later.) – Michael Anderson Dec 23 '11 at 08:57

3 Answers3