Re: WCHAR conversion problem
"I?aki" <Iaki@discussions.microsoft.com> wrote in message
news:E3DC8B26-374D-4000-A3BB-055B83C9B663@microsoft.com...
Hello everyone,
I need to convert a WCHAR[] to a char* but I don't know how can I do that.
I'm not c++ expert... and I'm using an API with WCHARs...
What I need to convert is fileName of the following struct to a char*:
typedef struct _FbwfCacheDetail {
ULONG cacheSize; // size of cache used by the file
ULONG openHandleCount; // number of currently opened handles
ULONG fileNameLength; // file name length in bytes
WCHAR fileName[1]; // file name (may not be
null-terminated)
} FbwfCacheDetail, *PFbwfCacheDetail;
I've tried the following:
char *FileName = (char *)malloc( MAX_FILENAME_LENGTH );
....
wcstombs_s(&wcharsConverted,FileName,MAX_FILENAME_LENGTH,Cache_detail->fileName,MAX_FILENAME_LENGTH);
where Cache_detail->fileName is suposed to contain fileName[], but
FileName
is empty while wcharsConverted seems to be OK.
Thanks in advance...
I've never used this function, preferring instead the Kernel function
"WideCharToMultiByte". For all it's faults, it has never failed for me with
standard settings. The problem wth wctombs_s is this difference between
"sizeInBytes" and "count". You have to read the help topic about 100 times
before the difference becomes clear as the description for the two variables
is essentially identical:
[in] sizeInBytes : Buffer size for mbstr
[in] count : The maximum number of bytes that can be stored in the multibyte
output string, or _TRUNCATE.
One would think that the size of the buffer would be the "maximum number" of
bytes that it can hold. Well, not so!
Ok, just kidding, they are the same, but why ask for the same info twice?
The rationalization for this discrepancy is as follows:
Line 1:
------
The count parameter indicates the maximum number of bytes that can be stored
in the multibyte output string (that is, the size of mbstr). In general, it
is not known how many bytes will be required when converting a
wide-character string. Some wide characters will require only one byte in
the output string; others require two. If there are two bytes in the
multibyte output string for every wide character in the input string
(including the wide character NULL), the result is guaranteed to fit.
This tells you that there is no difference [QUOTE]"that is, the size of
mbstr"[/QUOTE] , but offers an explanation for the two entries. So now it is
clear right?
Then you have to re-read the help another 10 times to discover that the
difference is not clear because why the long explanation? Should I allocate
2x the output size? Will it be null terminated? and so on.
Then again you have to read the help to try to understand why something that
was perfectly clear [QUOTE]"that is, the size of mbstr"[/QUOTE] is followed
up by an explanation about why it isn't what they just said it was.
But we know it it is actually, so GOTO line 1 just to be sure.
----------------------------------------
Well anyway, try using this simple macro and it should solve all your
problems:
// Returned _pmb must be freed by the caller:
#define WC2MB(_pmb,wide)
\
do
\
{
\
if(int iSize = WideCharToMultiByte(CP_ACP, 0, wide, (int)wcslen(wide),
NULL, 0, NULL, NULL)) \
{
\
_pmb = (char*)malloc(iSize+1);
\
WideCharToMultiByte(CP_ACP, 0, wide, (int)wcslen(wide), _pmb, iSize,
NULL, NULL); \
_pmb[iSize] = 0;
\
}
\
} while(false)
----------------------------------------
Never failed for me yet. Just don't forget to free the buffer!
- Alan Carre