How to convert char* to wchar_t*? - c++

I've tried implementing a function like this, but unfortunately it doesn't work:
const wchar_t *GetWC(const char *c)
{
const size_t cSize = strlen(c)+1;
wchar_t wc[cSize];
mbstowcs (wc, c, cSize);
return wc;
}
My main goal here is to be able to integrate normal char strings in a Unicode application. Any advice you guys can offer is greatly appreciated.

In your example, wc is a local variable which will be deallocated when the function call ends. This puts you into undefined behavior territory.
The simple fix is this:
const wchar_t *GetWC(const char *c)
{
const size_t cSize = strlen(c)+1;
wchar_t* wc = new wchar_t[cSize];
mbstowcs (wc, c, cSize);
return wc;
}
Note that the calling code will then have to deallocate this memory, otherwise you will have a memory leak.

Use a std::wstring instead of a C99 variable length array. The current standard guarantees a contiguous buffer for std::basic_string. E.g.,
std::wstring wc( cSize, L'#' );
mbstowcs( &wc[0], c, cSize );
C++ does not support C99 variable length arrays, and so if you compiled your code as pure C++, it would not even compile.
With that change your function return type should also be std::wstring.
Remember to set relevant locale in main.
E.g., setlocale( LC_ALL, "" ).

const char* text_char = "example of mbstowcs";
size_t length = strlen(text_char );
Example of usage "mbstowcs"
std::wstring text_wchar(length, L'#');
//#pragma warning (disable : 4996)
// Or add to the preprocessor: _CRT_SECURE_NO_WARNINGS
mbstowcs(&text_wchar[0], text_char , length);
Example of usage "mbstowcs_s"
Microsoft suggest to use "mbstowcs_s" instead of "mbstowcs".
Links:
Mbstowcs example
mbstowcs_s, _mbstowcs_s_l
wchar_t text_wchar[30];
mbstowcs_s(&length, text_wchar, text_char, length);

You're returning the address of a local variable allocated on the stack. When your function returns, the storage for all local variables (such as wc) is deallocated and is subject to being immediately overwritten by something else.
To fix this, you can pass the size of the buffer to GetWC, but then you've got pretty much the same interface as mbstowcs itself. Or, you could allocate a new buffer inside GetWC and return a pointer to that, leaving it up to the caller to deallocate the buffer.

I did something like this. The first 2 zeros are because I don't know what kind of ascii type things this command wants from me. The general feeling I had was to create a temp char array. pass in the wide char array. boom. it works. The +1 ensures that the null terminating character is in the right place.
char tempFilePath[MAX_PATH] = "I want to convert this to wide chars";
int len = strlen(tempFilePath);
// Converts the path to wide characters
int needed = MultiByteToWideChar(0, 0, tempFilePath, len + 1, strDestPath, len + 1);

Andrew Shepherd 's answer.
Andrew Shepherd 's answer is Good for me, I add up some fix :
1, remove the ending char L'\0', casue sometime it will trouble.
2, use mbstowcs_s
std::wstring wtos(std::string& value){
const size_t cSize = value.size() + 1;
std::wstring wc;
wc.resize(cSize);
size_t cSize1;
mbstowcs_s(&cSize1, (wchar_t*)&wc[0], cSize, value.c_str(), cSize);
wc.pop_back();
return wc;
}

The question has several problems, but so do some of the answers. The idea of returning a pointer to allocated memory "and leaving it up to the caller to de-allocate" is asking for trouble. As a rule the best pattern is always to allocate and de-allocate within the same function. For example, something like:
wchar_t* buffer = new wchar_t[get_wcb_size(str)];
mbstowcs(buffer, str, get_wcb_size(str) + 1);
...
delete[] buffer;
In general, this requires two functions, one the caller calls to find out how much memory to allocate and a second to initialize or fill the allocated memory.
Unfortunately, the basic idea of using a function to return a "new" object is problematic -- not inherently, but because of the C++ inheritance of C memory handling. Using C++ and STL's strings/wstrings/strstreams is a better solution, but I felt the memory allocation thing needed to be better addressed.

Your problem has nothing to do with encodings, it's a simple matter of understanding basic C++. You are returning a pointer to a local variable from your function, which will have gone out of scope by the time anyone can use it, thus creating undefined behaviour (i.e. a programming error).
Follow this Golden Rule: "If you are using naked char pointers, you're Doing It Wrong. (Except for when you aren't.)"
I've previously posted some code to do the conversion and communicating the input and output in C++ std::string and std::wstring objects.

auto Ascii_To_Wstring = [](int code)->std::wstring
{
if (code>255 || code<0 )
{
throw std::runtime_error("Incorrect ASCII code");
}
std::string s{ char(code) };
std::wstring w{ s.begin(),s.end() };
return w;
};

Related

Show glGetString output in message box [duplicate]

I've tried implementing a function like this, but unfortunately it doesn't work:
const wchar_t *GetWC(const char *c)
{
const size_t cSize = strlen(c)+1;
wchar_t wc[cSize];
mbstowcs (wc, c, cSize);
return wc;
}
My main goal here is to be able to integrate normal char strings in a Unicode application. Any advice you guys can offer is greatly appreciated.
In your example, wc is a local variable which will be deallocated when the function call ends. This puts you into undefined behavior territory.
The simple fix is this:
const wchar_t *GetWC(const char *c)
{
const size_t cSize = strlen(c)+1;
wchar_t* wc = new wchar_t[cSize];
mbstowcs (wc, c, cSize);
return wc;
}
Note that the calling code will then have to deallocate this memory, otherwise you will have a memory leak.
Use a std::wstring instead of a C99 variable length array. The current standard guarantees a contiguous buffer for std::basic_string. E.g.,
std::wstring wc( cSize, L'#' );
mbstowcs( &wc[0], c, cSize );
C++ does not support C99 variable length arrays, and so if you compiled your code as pure C++, it would not even compile.
With that change your function return type should also be std::wstring.
Remember to set relevant locale in main.
E.g., setlocale( LC_ALL, "" ).
const char* text_char = "example of mbstowcs";
size_t length = strlen(text_char );
Example of usage "mbstowcs"
std::wstring text_wchar(length, L'#');
//#pragma warning (disable : 4996)
// Or add to the preprocessor: _CRT_SECURE_NO_WARNINGS
mbstowcs(&text_wchar[0], text_char , length);
Example of usage "mbstowcs_s"
Microsoft suggest to use "mbstowcs_s" instead of "mbstowcs".
Links:
Mbstowcs example
mbstowcs_s, _mbstowcs_s_l
wchar_t text_wchar[30];
mbstowcs_s(&length, text_wchar, text_char, length);
You're returning the address of a local variable allocated on the stack. When your function returns, the storage for all local variables (such as wc) is deallocated and is subject to being immediately overwritten by something else.
To fix this, you can pass the size of the buffer to GetWC, but then you've got pretty much the same interface as mbstowcs itself. Or, you could allocate a new buffer inside GetWC and return a pointer to that, leaving it up to the caller to deallocate the buffer.
Andrew Shepherd 's answer.
Andrew Shepherd 's answer is Good for me, I add up some fix :
1, remove the ending char L'\0', casue sometime it will trouble.
2, use mbstowcs_s
std::wstring wtos(std::string& value){
const size_t cSize = value.size() + 1;
std::wstring wc;
wc.resize(cSize);
size_t cSize1;
mbstowcs_s(&cSize1, (wchar_t*)&wc[0], cSize, value.c_str(), cSize);
wc.pop_back();
return wc;
}
The question has several problems, but so do some of the answers. The idea of returning a pointer to allocated memory "and leaving it up to the caller to de-allocate" is asking for trouble. As a rule the best pattern is always to allocate and de-allocate within the same function. For example, something like:
wchar_t* buffer = new wchar_t[get_wcb_size(str)];
mbstowcs(buffer, str, get_wcb_size(str) + 1);
...
delete[] buffer;
In general, this requires two functions, one the caller calls to find out how much memory to allocate and a second to initialize or fill the allocated memory.
Unfortunately, the basic idea of using a function to return a "new" object is problematic -- not inherently, but because of the C++ inheritance of C memory handling. Using C++ and STL's strings/wstrings/strstreams is a better solution, but I felt the memory allocation thing needed to be better addressed.
Your problem has nothing to do with encodings, it's a simple matter of understanding basic C++. You are returning a pointer to a local variable from your function, which will have gone out of scope by the time anyone can use it, thus creating undefined behaviour (i.e. a programming error).
Follow this Golden Rule: "If you are using naked char pointers, you're Doing It Wrong. (Except for when you aren't.)"
I've previously posted some code to do the conversion and communicating the input and output in C++ std::string and std::wstring objects.
I did something like this. The first 2 zeros are because I don't know what kind of ascii type things this command wants from me. The general feeling I had was to create a temp char array. pass in the wide char array. boom. it works. The +1 ensures that the null terminating character is in the right place.
char tempFilePath[MAX_PATH] = "I want to convert this to wide chars";
int len = strlen(tempFilePath);
// Converts the path to wide characters
int needed = MultiByteToWideChar(0, 0, tempFilePath, len + 1, strDestPath, len + 1);
auto Ascii_To_Wstring = [](int code)->std::wstring
{
if (code>255 || code<0 )
{
throw std::runtime_error("Incorrect ASCII code");
}
std::string s{ char(code) };
std::wstring w{ s.begin(),s.end() };
return w;
};

C string to wide C string assignment

I'm a little confused about C strings and wide C strings. For the sake of this question, assume that I using Microsoft Visual Studio 2010 Professional. Please let me know if any of my information is incorrect.
I have a struct with a const wchar_t* member which is used to store a name.
struct A
{
const wchar_t* name;
};
When I assign object 'a' a name as so:
int main()
{
A a;
const wchar_t* w_name = L"Tom";
a.name = w_name;
return 0;
}
That is just copying the memory address that w_name points to into a.name. Now w_name and a.name are both wide character pointers which point to the same address in memory.
If I am correct, then I am wondering what to do about a situation like this. I am reading in a C string from an XML attribute using tinyxml2.
tinyxml2::XMLElement* pElement;
// ...
const char* name = pElement->Attribute("name");
After I have my C string, I am converting it to a wide character string as follows:
size_t newsize = strlen(name) + 1;
wchar_t * wcName = new wchar_t[newsize];
size_t convertedChars = 0;
mbstowcs_s(&convertedChars, wcName, newsize, name, _TRUNCATE);
a.name = wcName;
delete[] wcName;
If I am correct so far, then the line:
a.name = wcName;
is just copying the memory address of the first character of array wcName into a.name. However, I am deleting wcName directly after assigning this pointer which would make it point to garbage.
How can I convert my C string into a wide character C string and then assign it to a.name?
The easiest approach is probably to task you name variable with the management of the memory. This, in turn, is easily done by declaring it as
std::wstring name;
These guys don't have a concept of independent content and object mutation, i.e., you can't really make the individual characters const and making the entire object const would prevent it from being assigned to.
You can do this while using a std::wstring without relying on the additional temporary conversion buffer allocation and destruction. Not tremendously important unless you're overtly concerned about heap fragmentation or on a limited system (aka Windows Phone). It just takes a little setup on the front side. Let the standard library manage the memory for you (with a little nudge).
class A
{
...
std::wstring a;
};
// Convert the string (I'm assuming it is UTF8) to wide char
int wlen = MultiByteToWideChar(CP_UTF8, 0, name, -1, NULL, NULL);
if (wlen > 0)
{
// reserve space. std::wstring gives us the terminator slot
// for free, so don't include that. MB2WC above returns the
// length *including* the terminator.
a.resize(wlen-1);
MultiByteToWideChar(CP_UTF8, 0, name, -1, &a[0], wlen);
}
else
{ // no conversion available/possible.
a.clear();
}
On a complete side-note, you can build TinyXML to use the standard library and std::string rather than char *, which doesn't really help you much here, but may save you a ton of future strlen() calls later on.
As you correctly mentioned a.name is just a pointer which doesn't suppose any allocated string storage. You must manage it manually using new or static/scoped array.
To get rid of these boring things just use one of available string classes: CStringW from ATL (easy to use but MS-specific) or std::wstring from STL (C++ standard, but not so easy to convert from char*):
#include <atlstr.h>
// Conversion ANSI -> Wide is automatic
const CStringW name(pElement->Attribute("name"));
Unfortunately, std::wstring usage with char* is not so easy.
See conversion functon here: How to convert std::string to LPCWSTR in C++ (Unicode)

Why does windows need the size when calling a function?

I am trying to learn a little c++ and I have a silly question. Consider this code:
TCHAR tempPath[255];
GetTempPath(255, tempPath);
Why does windows need the size of the var tempPath? I see that the GetTempPath is declared something like:
GetTempPath(dword size, buf LPTSTR);
How can windows change the buf value without the & operator? Should not the function be like that?
GetTempPath(buf &LPTSTR);
Can somebody provide a simple GetTempPath implementation sample so I can see how size is used?
EDIT:
Thanks for all your answers, they are all correct and I gave you all +1. But what I meant by "Can somebody provide a simple GetTempPath implementation) is that i have tried to code a function similar to the one windows uses, as follow:
void MyGetTempPath(int size, char* buf)
{
buf = "C:\\test\\";
}
int main(int argc, char *argv[])
{
char* tempPath = new TCHAR[255];
GetTempPathA(255, tempPath);
MessageBoxA(0, tempPath, "test", MB_OK);
return EXIT_SUCCESS;
}
But it does not work. MessageBox displays a "##$' string. How should MyGetTempPath be coded to work properly?
Windows needs the size as a safety precaution. It could crash the application if it copies characters past the end of the buffer. When you supply the length, it can prevent that.
Array variables work like pointers. They point to the data in the array. So there is no need for the & operator.
Not sure what kind of example you are looking for. Like I said, it just needs to verify it doesn't write more characters than there's room for.
An array cannot be passed into functions by-value. Instead, it's converted to a pointer to the first element, and that's passed to the function. Having a (non-const) pointer to data allows modification:
void foo(int* i)
{
if (i) (don't dereference null)
*i = 5; // dereference pointer, modify int
}
Likewise, the function now has a pointer to a TCHAR it can write to. It takes the size, then, so it knows exactly how many TCHAR's exist after that initial one. Otherwise it wouldn't know how large the array is.
GetTempPath() outputs into your "tempPath" character array. If you don't tell it how much space there is allocated in the array (255), it has no way of knowing whether or not it will have enough room to write the path string into tempPath.
Character arrays in C/C++ are pretty much just pointers to locations in memory. They don't contain other information about themselves, like instances of C++ or Java classes might. The meat and potatoes of the Windows API was designed before C++ really had much inertia, I think, so you'll often have to use older C style techniques and built-in data types to work with it.
Following wrapper can be tried, if you want to avoid the size:
template<typename CHAR_TYPE, unsigned int SIZE>
void MyGetTempPath (CHAR_TYPE (&array)[SIZE]) // 'return' value can be your choice
{
GetTempPath(SIZE, array);
}
Now you can use like below:
TCHAR tempPath[255];
MyGetTempPath(tempPath); // No need to pass size, it will count automatically
In your other question, why we do NOT use following:
GetTempPath(buf &LPTSTR);
is because, & is used when you want to pass a data type by reference (not address). I am not aware what buf is typecasted to but it should be some pointer type.
Can somebody provide a simple
GetTempPath implementation sample so I
can see how size is used?
First way (based on MAX_PATH constant):
TCHAR szPath[MAX_PATH];
GetTempPath(MAX_PATH, szPath);
Second way (based on GetTempPath description):
DWORD size;
LPTSTR lpszPath;
size = GetTempPath(0, NULL);
lpszPath = new TCHAR[size];
GetTempPath(size, lpszPath);
/* some code here */
delete[] lpszPath;
How can windows change the buf value without the & operator?
& operator is not needed because array name is the pointer to first array element (or to all array). Try next code to demonstrate this:
TCHAR sz[1];
if ((void*)sz == (void*)&sz) _tprintf(TEXT("sz equals to &sz \n"));
if ((void*)sz == (void*)&(sz[0])) _tprintf(TEXT("sz equals to &(sz[0]) \n"));
As requested, a very simple implementation.
bool MyGetTempPath(size_t size, char* buf)
{
const char* path = "C:\\test\\";
size_t len = strlen(path);
if(buf == NULL)
return false;
if(size < len + 1)
return false;
strncpy(buf, path, size);
return true;
}
An example call to the new function:
char buffer[256];
bool success = MyGetTempPath(256, buffer);
from http://msdn.microsoft.com/en-us/library/aa364992(v=vs.85).aspx
DWORD WINAPI GetTempPath(
__in DWORD nBufferLength,
__out LPTSTR lpBuffer
);
so GetTempPath is defined something like
GetTempPath(DWORD nBufferLength, LPTSTR& lpBuffer);
What mean, that compiler passes the value lpBuffer by referenece.

CString : What does (TCHAR*)(this + 1) mean?

In the CString header file (be it Microsoft's or Open Foundation Classes - http://www.koders.com/cpp/fid035C2F57DD64DBF54840B7C00EA7105DFDAA0EBD.aspx#L77 ), there is the following code snippet
struct CStringData
{
long nRefs;
int nDataLength;
int nAllocLength;
TCHAR* data() { return (TCHAR*)(&this[1]); };
...
};
What does the (TCHAR*)(&this[1]) indicate?
The CStringData struct is used in the CString class (http :// www.koders.com/cpp/fid100CC41B9D5E1056ED98FA36228968320362C4C1.aspx).
Any help is appreciated.
CString has lots of internal tricks which make it look like a normal string when passed e.g. to printf functions, despite actually being a class - without having to cast it to LPCTSTR in the argument list, e.g., in the case of varargs (...) in e.g. a printf. Thus trying to understand a single individual trick or function in the CString implementation is bad news. (The data function is an internal function which gets the 'real' buffer associated with the string.)
There's a book, MFC Internals that goes into it, and IIRC the Blaszczak book might touch it.
EDIT: As for what the expression actually translates to in terms of raw C++:-
TCHAR* data() { return (TCHAR*)(&this[1]); };
this says "pretend you're actually the first entry in an array of items allocated together. Now, the second item isnt actually a CString, it's a normal NUL terminated buffer of either Unicode or normal characters - i.e., an LPTSTR".
Another way of expressing the same thing is:
TCHAR* data() { return (TCHAR*)(this + 1); };
When you add 1 to a pointer to T, you actually add 1* sizeof T in terms of a raw memory address. So if one has a CString located at 0x00000010 with sizeof(CString) = 4, data will return a pointer to a NUL terminated array of chars buffer starting at 0x00000014
But just understanding this one thing out of context isnt necessarily a good idea.
Why do you need to know?
It returns the memory area that is immediately after the CStringData structure as an array of TCHAR characters.
You can understand why they are doing this if you look at the CString.cpp file:
static const struct {
CStringData data;
TCHAR ch;
} str_empty = {{-1, 0, 0}, 0};
CStringData* pData = (CStringData*)mem_alloc(sizeof(CStringData) + size*sizeof(TCHAR));
They do this trick, so that CString looks like a normal data buffer, and when you ask for the getdata it skips the CStringData structure and points directly to the real data buffer like char*

Can I get a non-const C string back from a C++ string?

Const-correctness in C++ is still giving me headaches. In working with some old C code, I find myself needing to assign turn a C++ string object into a C string and assign it to a variable. However, the variable is a char * and c_str() returns a const char []. Is there a good way to get around this without having to roll my own function to do it?
edit: I am also trying to avoid calling new. I will gladly trade slightly more complicated code for less memory leaks.
C++17 and newer:
foo(s.data(), s.size());
C++11, C++14:
foo(&s[0], s.size());
However this needs a note of caution: The result of &s[0]/s.data()/s.c_str() is only guaranteed to be valid until any member function is invoked that might change the string. So you should not store the result of these operations anywhere. The safest is to be done with them at the end of the full expression, as my examples do.
Pre C++-11 answer:
Since for to me inexplicable reasons nobody answered this the way I do now, and since other questions are now being closed pointing to this one, I'll add this here, even though coming a year too late will mean that it hangs at the very bottom of the pile...
With C++03, std::string isn't guaranteed to store its characters in a contiguous piece of memory, and the result of c_str() doesn't need to point to the string's internal buffer, so the only way guaranteed to work is this:
std::vector<char> buffer(s.begin(), s.end());
foo(&buffer[0], buffer.size());
s.assign(buffer.begin(), buffer.end());
This is no longer true in C++11.
There is an important distinction you need to make here: is the char* to which you wish to assign this "morally constant"? That is, is casting away const-ness just a technicality, and you really will still treat the string as a const? In that case, you can use a cast - either C-style or a C++-style const_cast. As long as you (and anyone else who ever maintains this code) have the discipline to treat that char* as a const char*, you'll be fine, but the compiler will no longer be watching your back, so if you ever treat it as a non-const you may be modifying a buffer that something else in your code relies upon.
If your char* is going to be treated as non-const, and you intend to modify what it points to, you must copy the returned string, not cast away its const-ness.
I guess there is always strcpy.
Or use char* strings in the parts of your C++ code that must interface with the old stuff.
Or refactor the existing code to compile with the C++ compiler and then to use std:string.
There's always const_cast...
std::string s("hello world");
char *p = const_cast<char *>(s.c_str());
Of course, that's basically subverting the type system, but sometimes it's necessary when integrating with older code.
If you can afford extra allocation, instead of a recommended strcpy I would consider using std::vector<char> like this:
// suppose you have your string:
std::string some_string("hello world");
// you can make a vector from it like this:
std::vector<char> some_buffer(some_string.begin(), some_string.end());
// suppose your C function is declared like this:
// some_c_function(char *buffer);
// you can just pass this vector to it like this:
some_c_function(&some_buffer[0]);
// if that function wants a buffer size as well,
// just give it some_buffer.size()
To me this is a bit more of a C++ way than strcpy. Take a look at Meyers' Effective STL Item 16 for a much nicer explanation than I could ever provide.
You can use the copy method:
len = myStr.copy(cStr, myStr.length());
cStr[len] = '\0';
Where myStr is your C++ string and cStr a char * with at least myStr.length()+1 size. Also, len is of type size_t and is needed, because copy doesn't null-terminate cStr.
Just use const_cast<char*>(str.data())
Do not feel bad or weird about it, it's perfectly good style to do this.
It's guaranteed to work in C++11. The fact that it's const qualified at all is arguably a mistake by the original standard before it; in C++03 it was possible to implement string as a discontinuous list of memory, but no one ever did it. There is not a compiler on earth that implements string as anything other than a contiguous block of memory, so feel free to treat it as such with complete confidence.
If you know that the std::string is not going to change, a C-style cast will work.
std::string s("hello");
char *p = (char *)s.c_str();
Of course, p is pointing to some buffer managed by the std::string. If the std::string goes out of scope or the buffer is changed (i.e., written to), p will probably be invalid.
The safest thing to do would be to copy the string if refactoring the code is out of the question.
std::string vString;
vString.resize(256); // allocate some space, up to you
char* vStringPtr(&vString.front());
// assign the value to the string (by using a function that copies the value).
// don't exceed vString.size() here!
// now make sure you erase the extra capacity after the first encountered \0.
vString.erase(std::find(vString.begin(), vString.end(), 0), vString.end());
// and here you have the C++ string with the proper value and bounds.
This is how you turn a C++ string to a C string. But make sure you know what you're doing, as it's really easy to step out of bounds using raw string functions. There are moments when this is necessary.
If c_str() is returning to you a copy of the string object internal buffer, you can just use const_cast<>.
However, if c_str() is giving you direct access tot he string object internal buffer, make an explicit copy, instead of removing the const.
Since c_str() gives you direct const access to the data structure, you probably shouldn't cast it. The simplest way to do it without having to preallocate a buffer is to just use strdup.
char* tmpptr;
tmpptr = strdup(myStringVar.c_str();
oldfunction(tmpptr);
free tmpptr;
It's quick, easy, and correct.
In CPP, if you want a char * from a string.c_str()
(to give it for example to a function that only takes a char *),
you can cast it to char * directly to lose the const from .c_str()
Example:
launchGame((char *) string.c_str());
C++17 adds a char* string::data() noexcept overload. So if your string object isn't const, the pointer returned by data() isn't either and you can use that.
Is it really that difficult to do yourself?
#include <string>
#include <cstring>
char *convert(std::string str)
{
size_t len = str.length();
char *buf = new char[len + 1];
memcpy(buf, str.data(), len);
buf[len] = '\0';
return buf;
}
char *convert(std::string str, char *buf, size_t len)
{
memcpy(buf, str.data(), len - 1);
buf[len - 1] = '\0';
return buf;
}
// A crazy template solution to avoid passing in the array length
// but loses the ability to pass in a dynamically allocated buffer
template <size_t len>
char *convert(std::string str, char (&buf)[len])
{
memcpy(buf, str.data(), len - 1);
buf[len - 1] = '\0';
return buf;
}
Usage:
std::string str = "Hello";
// Use buffer we've allocated
char buf[10];
convert(str, buf);
// Use buffer allocated for us
char *buf = convert(str);
delete [] buf;
// Use dynamic buffer of known length
buf = new char[10];
convert(str, buf, 10);
delete [] buf;