How to convert char* to wchar_t*?

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.

How to convert char* to wchar_t*?

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