如何在AURenderCallback中将两个Audio Unit合并到AudioBufferList

首先要明确你需要的是音频混音（叠加播放）还是音频拼接（先后播放），两种场景的实现逻辑不同，以下是具体实现：

前提确认 #

• 确保buffList1、buffList2与ioData的音频格式完全一致（采样率、位深、通道数、字节序）
• 保证ioData的缓冲区有足够空间容纳合并后的数据，避免内存越界

场景1：音频混音（两个音频叠加） #

如果需要同时播放两个音频，将波形叠加，需逐样本相加并处理溢出：

32位浮点PCM（iOS/macOS常见格式）

OSStatus PlayCallback(void *inRefCon,
                        AudioUnitRenderActionFlags *ioActionFlags,
                        const AudioTimeStamp *inTimeStamp,
                        UInt32 inBusNumber,
                        UInt32 inNumberFrames,
                        AudioBufferList *ioData) {
    ALNPlayer *player = (__bridge ALNPlayer *)inRefCon;
    
    OSStatus status;
    
    // 读取音频1数据到buffList1，检查错误
    status = AudioConverterFillComplexBuffer(player->audioConverter1, lyInInputDataProcV1, inRefCon, &inNumberFrames, player->buffList1, NULL);
    if (status != noErr) return status;
    
    // 读取音频2数据到buffList2，检查错误
    status = AudioConverterFillComplexBuffer(player->audioConverter2, lyInInputDataProcV2, inRefCon, &inNumberFrames, player->buffList2, NULL);
    if (status != noErr) return status;
    
    // 转换为浮点指针（单通道场景）
    float *ioPtr = (float *)ioData->mBuffers[0].mData;
    float *buff1Ptr = (float *)player->buffList1->mBuffers[0].mData;
    float *buff2Ptr = (float *)player->buffList2->mBuffers[0].mData;
    
    // 逐样本叠加，限幅防止失真（浮点范围[-1.0, 1.0]）
    for (UInt32 i = 0; i < inNumberFrames; i++) {
        float mixed = buff1Ptr[i] + buff2Ptr[i];
        ioPtr[i] = fmaxf(fminf(mixed, 1.0f), -1.0f);
    }
    
    // 更新ioData的字节数
    ioData->mBuffers[0].mDataByteSize = inNumberFrames * sizeof(float);
    
    return noErr;
}

16位整型PCM

如果是16位整数格式，需用32位整型中转避免溢出：

// 转换为16位整型指针
SInt16 *ioPtr = (SInt16 *)ioData->mBuffers[0].mData;
SInt16 *buff1Ptr = (SInt16 *)player->buffList1->mBuffers[0].mData;
SInt16 *buff2Ptr = (SInt16 *)player->buffList2->mBuffers[0].mData;

for (UInt32 i = 0; i < inNumberFrames; i++) {
    SInt32 sum = (SInt32)buff1Ptr[i] + (SInt32)buff2Ptr[i];
    // 限幅到16位范围[-32768, 32767]
    sum = fmaxf(fminf(sum, 32767), -32768);
    ioPtr[i] = (SInt16)sum;
}

ioData->mBuffers[0].mDataByteSize = inNumberFrames * sizeof(SInt16);

场景2：音频拼接（先播放buffList1，再播放buffList2） #

如果需要按顺序播放两个音频，直接拼接缓冲区即可：

OSStatus PlayCallback(void *inRefCon,
                        AudioUnitRenderActionFlags *ioActionFlags,
                        const AudioTimeStamp *inTimeStamp,
                        UInt32 inBusNumber,
                        UInt32 inNumberFrames,
                        AudioBufferList *ioData) {
    ALNPlayer *player = (__bridge ALNPlayer *)inRefCon;
    
    OSStatus status;
    
    // 读取音频1数据到buffList1
    status = AudioConverterFillComplexBuffer(player->audioConverter1, lyInInputDataProcV1, inRefCon, &inNumberFrames, player->buffList1, NULL);
    if (status != noErr) return status;
    
    // 读取音频2数据到buffList2
    status = AudioConverterFillComplexBuffer(player->audioConverter2, lyInInputDataProcV2, inRefCon, &inNumberFrames, player->buffList2, NULL);
    if (status != noErr) return status;
    
    UInt32 buff1Size = player->buffList1->mBuffers[0].mDataByteSize;
    UInt32 buff2Size = player->buffList2->mBuffers[0].mDataByteSize;
    
    // 先复制buffList1到ioData起始位置
    memcpy(ioData->mBuffers[0].mData, player->buffList1->mBuffers[0].mData, buff1Size);
    // 再复制buffList2到buffList1的末尾
    void *offsetPtr = (uint8_t *)ioData->mBuffers[0].mData + buff1Size;
    memcpy(offsetPtr, player->buffList2->mBuffers[0].mData, buff2Size);
    
    // 更新ioData的总字节数
    ioData->mBuffers[0].mDataByteSize = buff1Size + buff2Size;
    
    return noErr;
}

额外注意事项 #

• 多通道场景：如果是立体声或多通道，需要遍历AudioBufferList中的每个mBuffers元素，分别处理每个通道的数据
• 缓冲区大小：拼接场景下，要确保ioData的缓冲区大小≥buff1Size + buff2Size，否则会触发内存越界
• 错误处理：每次调用AudioConverterFillComplexBuffer后必须检查返回状态，避免处理无效数据

内容的提问来源于stack exchange，提问作者Alan Luo