How to convert wchar_t** to char**?

I get argv as wchar_t** (see below), because I need to work with unicode, but I need to convert it to 开发者_如何转开发char **. How can I do that?

int wmain(int argc, wchar_t** argv) 
{ 

There is more than one way to do this. Depending on your environment and available compiler/standard library/other libraries, you have at least three choices:

1. Use std::locale and std::codecvt<> facet;
2. use C locale functions like std::mbstowcs();
3. use 3rd party functions like iconv() on *nix or WideCharToMultiByte() on Windows.

Do you really need to do the conversion?

You should realise that often (especially on Windows) the conversion from wchar_t string to char string is a lossy conversion. The character set used by system for char strings is often not UTF-8. E.g. if you convert a file name with national characters or in some Asian language to char string you are most likely going to get something that will not be really usable to access the original file.

This does the trick:

#define MAXLEN 512
   char tx[MAXLEN];
   mbstowcs(argv[i], tx, MAXLEN);

Why do you need to convert? In most of the cases you need to change your project settings, so everything accepts wide chars. If a third party library wants non-Unicode string, then you need to recompile it with the proper options for Unicode. If there are no proper options for Unicode, I would get rid of this library and find (write) a better one.

I needed this for a hack to get ancient code working. It is not thread safe because it is returning a pointer to an internal static array. @red-soft-adair answer converts in the opposite direction.

char *WideStringToCString(wchar_t *wcStr)
{
#define MAXLEN 512

    static char tx[MAXLEN];
    wcstombs(tx, wcStr, MAXLEN-1);
    tx[MAXLEN - 1] = 0;
    return tx;
}