Is there way to define return type inside the body of function in C++?

Question

How i can define return type inside function at run-time? I have a member char* m_data; and I want to return a conversion of m_data into different types in different cases.

?type? getData() const
{
    switch(this->p_header->WAVE_F.bitsPerSample)
    {
        case 8:
        {
            // return type const char *
            break;
        }
        case 16:
        {
            // return type const short *
            break;
        }
        case 32:
        {
            // return type const int *
            break;
        }
    }
}

Answer 1

No, but as you always return a pointer you could just return a void*. Beware the caller would have no option to figure out what is behind the pointer, so you better try to wrap the return value in boost::variant<char*,short*,int*> or boost::any/cdiggins::any

Answer 2

Make a getter for bitsPerSample and let the caller choose one of the appropriate methods:

int getBitsPerSample(){
    return p_header->WAVE_F.bitsPerSample;
}

const char* getDataAsCharPtr() {
    // return type const char *
}

const short* getDataAsShortPtr() {
    // return type const short *
}

const int* getDataAsIntPtr() {
    // return type const int *
}

Answer 3

Not directly, I suggest to use something like :

const char* getDataAsChar() const
{
    if (this->p_header->WAVE_F.bitsPerSample != 8) return nullptr;
    //return data as const char*
}

const short* getDataAsShort() const
{
    if (this->p_header->WAVE_F.bitsPerSample != 16) return nullptr;
    //return data as const short*
}

const int* getDataAsInt() const
{
    if (this->p_header->WAVE_F.bitsPerSample != 32) return nullptr;
    //return data as const int*
}