What is the proper way to convert an int to wchar_t?

Question

Say you have "methods"

int Read(...)
{
    unsigned char Byte = 0;
    if(!ReadFile(.., &byte, 1,...))
        return -1;

    return Byte;
}


int ReadBlock(LPWSTR buffer, int cchBuffer ...)
{
    int c = 0;
    int cnt = 0;
    do
    {
        if( (c=Read(...)) != -1 )
            buffer[num++] = c; // Here.
    } while( num < ccBuffer );
    return cnt;
}

What is the proper way to get that int correctly to WCHAR?

Answer 1

After reading When should static_cast, dynamic_cast and reinterpret_cast be used?, I realize it was my lack of knowledge about casting that triggered me to ask this question.

Answer 2

Use mbstowcs (multibyte string to wide character string) :

int ReadBlock(LPWSTR buffer, int cchBuffer ...)
{
    int c = 0;
    std::vector<char> narrow;
    while((c=Read(...)) != -1 )
       narrow.push_back(c);
    }
    narrow.push_back(0);
    mbstowcs(buffer, &narrow.front(), cchBuffer);
}

mbstowcs uses the current locale, so that should match the encoding of your input.

Answer 3

convert char <= => wchar
in windows:
MultiByteToWideChar
WideCharToMultiByte

in linux:
mbsrtowcs
wcsrtombs

Answer 4

#include<tchar.h>

int main()
{
    int integer = 0;
    wchar_t wideCharacter = (wchar_t)integer;

    return 0;

}