如何将数据缓冲区从 SDL_LoadWAV 复制到 SDL_AudioCVT？

Question

此函数应加载 wav 并将其转换为所需的格式。 我的问题是我不知道如何正确地将数据复制到 cvt。 我从文档中收集到的是音频数据是一个指向指针的 Uint8 指针。 我知道它必须指向某种数组，但我不知道它是如何设置的。

void load_audio()
{
    string full_path = default_path;
    full_path += path;
    if (wav_length > 0)
    {
        SDL_FreeWAV(wav_buffer);
    }
    if (SDL_LoadWAV(full_path.c_str(), &wav_spec, &wav_buffer, &wav_length) == NULL)
    {
        fprintf(stderr, "Could not open wav: %s\n", SDL_GetError());
        return;
    }
    // convert input format to U8 mono 32kHz
    SDL_AudioCVT cvt;
    SDL_BuildAudioCVT(&cvt, wav_spec.format, wav_spec.channels, wav_spec.freq, AUDIO_U8, 1, 32000);
    SDL_assert(cvt.needed);
    cvt.len = wav_length * wav_spec.channels * 2;  // length of original data * 2 for 16bit
    cvt.buf = (Uint8 *) SDL_malloc(cvt.len * cvt.len_mult);

    // How do I copy the data from wav_buffer to cvt.buf ???

    SDL_ConvertAudio(&cvt); // convert
}

wav 变量如下

SDL_AudioSpec wav_spec;
Uint32 wav_length;
Uint8 *wav_buffer;

旁注：我将加载的 wav 文件将始终是 16 位

Answer 1

答案是 memcpy() 和 cvt_len var 而不是 cvt.len

// convert input format to U8 mono 32kHz
SDL_AudioCVT cvt;
SDL_BuildAudioCVT(&cvt, wav_spec.format, wav_spec.channels, wav_spec.freq, AUDIO_U8, 1, 32000);
SDL_assert(cvt.needed); // obviously, this one is always needed.
cvt.len = wav_length * wav_spec.channels * 2;  // length of original data * 2 for 16bit
cvt.buf = (Uint8 *) SDL_malloc(cvt.len * cvt.len_mult); // allocate buffer
memcpy(cvt.buf, wav_buffer, wav_length); // copy wav data to cvt buffer
SDL_ConvertAudio(&cvt); // convert
wav_length = cvt.len_cvt; // len_cvt is the actaul length after conversion
memcpy(wav_buffer, cvt.buf, cvt.len_cvt); // copy converted cvt buffer back to wav buffer