c++ - 为什么 QueryInterface() 返回 E

问题描述

我正在编写一个项目，我遇到了一个QueryInterface()正在返回的错误E_INVALIDARG，我不知道哪个参数无效，因为两个参数都很简单，我是否忽略了一些简单的东西？

失败的那一点来自微软自己的特别文档，如果我做的事情完全错误，请告诉我

HRESULT CreateSecondaryBuffer(LPDIRECTSOUND8 lpDirectSound, LPDIRECTSOUNDBUFFER8* ppDsb8, LPCWSTR lpName) {
    WAVEFORMATEX wfx;
    DSBUFFERDESC dsbdesc;
    LPDIRECTSOUNDBUFFER pDsb = NULL;
    HRESULT hr = 0;

    void* lpRes;
    HRSRC hResInfo;
    HGLOBAL hRes;

    LPVOID lpvWrite;
    DWORD  dwLength;

    waveFile myWave;
    memset(&myWave, 0, sizeof(waveFile));
    // Find the WAVE resource. 

    hResInfo = FindResource(wc.hInstance, lpName, L"WAVE");
    if (hResInfo == NULL)
        return FALSE;

    // Load the WAVE resource. 

    hRes = LoadResource(wc.hInstance, hResInfo);
    if (hRes == NULL)
        return FALSE;

    // Lock the WAVE resource and play it. 

    lpRes = LockResource(hRes);
    char* wavMem = static_cast<char*>(lpRes);
    memcpy(&myWave, wavMem, 36);
    wavMem += 36;
    

    while (myWave.Subchunk2ID[0] != 'd' && myWave.Subchunk2ID[1] != 'a' && myWave.Subchunk2ID[2] != 't' && myWave.Subchunk2ID[3] != 'a') {
        memcpy(&myWave.Subchunk2ID, wavMem, 8);
        if (myWave.Subchunk2ID[0] != 'd' && myWave.Subchunk2ID[1] != 'a' && myWave.Subchunk2ID[2] != 't' && myWave.Subchunk2ID[3] != 'a') {
            wavMem += myWave.Subchunk2Size;
            wavMem += 8;
        }
    }
    wavMem += 8;

    char* soundData = new char[myWave.Subchunk2Size];
    memcpy(soundData, wavMem, myWave.Subchunk2Size);
    FreeResource(hRes);

    // Set up WAV format structure. 

    memset(&wfx, 0, sizeof(WAVEFORMATEX));
    wfx.wFormatTag = WAVE_FORMAT_PCM;
    wfx.nChannels = myWave.NumChannels;
    wfx.nSamplesPerSec = myWave.SampleRate;
    wfx.nBlockAlign = myWave.BlockAlign;
    wfx.nAvgBytesPerSec = wfx.nSamplesPerSec * wfx.nBlockAlign;
    wfx.wBitsPerSample = myWave.BitsPerSample;

    // Set up DSBUFFERDESC structure. 

    memset(&dsbdesc, 0, sizeof(DSBUFFERDESC));
    dsbdesc.dwSize = sizeof(DSBUFFERDESC);
    dsbdesc.dwFlags =
         DSBCAPS_CTRLVOLUME | DSBCAPS_GLOBALFOCUS;
    dsbdesc.dwBufferBytes = myWave.Subchunk2Size;
    dsbdesc.lpwfxFormat = &wfx;

    hr = lpDirectSound->CreateSoundBuffer(&dsbdesc, &pDsb, NULL);
    if (SUCCEEDED(hr))
    {
        if (DS_OK == pDsb->Lock(
            0,          // Offset at which to start lock.
            0,          // Size of lock; ignored because of flag.
            &lpvWrite,  // Gets address of first part of lock.
            &dwLength,  // Gets size of first part of lock.
            NULL,       // Address of wraparound not needed. 
            NULL,       // Size of wraparound not needed.
            DSBLOCK_ENTIREBUFFER))  // Flag.
        {
            memcpy(lpvWrite, soundData, dwLength);
            pDsb->Unlock(
                lpvWrite,   // Address of lock start.
                dwLength,   // Size of lock.
                NULL,       // No wraparound portion.
                0);         // No wraparound size.
        }
        hr = pDsb->QueryInterface(IID_IDirectSoundBuffer8, (LPVOID*)ppDsb8);
        pDsb->Release();
    }

    delete[] soundData;
    return hr;
}

标签： c++winapicomdirectsound

解决方案

只有 2 个参数可以QueryInterface()抱怨E_INVALIDARG，而且在你的CreateSecondaryBuffer()函数内部对我来说它们看起来很好。第一个参数被传递一个硬编码的 IID，所以问题必须出在第二个参数上，它期望被传递一个指针（在这种情况下）一个可以写入对象内存地址的IDirectSoundBuffer8*变量的指针。所以很可能，你的函数的参数没有像你期望的那样被传递给一个变量的有效指针。QueryInterface()DirectSoundBuffer8ppDsb8CreateSecondaryBuffer()IDirectSoundBuffer8*

更新：看到您添加的评论后，您确实将无效指针传递ppDsb8给CreateSecondaryBuffer(). 调用者的pMainSecondaryBuffer指针被错误地声明，并且在传递给时未初始化CreateSecondaryBuffer()，因此您将错误的指针传递给QueryInterface()，因此出现E_INVALIDARG错误。

改用这个：

LPDIRECTSOUND8 lpDevice = ...; // <-- make sure it points to a valid device object...
LPDIRECTSOUNDBUFFER8 pMainSecondaryBuffer;
CreateSecondaryBuffer(lpDevice, &pMainSecondaryBuffer, L"mainSong");

c++ - 为什么 QueryInterface() 返回 E_INVALIDARG？

问题描述

解决方案

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