ALSA的snd_pcm_writei在阻止模式下的行为

Question

我稍微修改了从ALSA Project website获取的演示，以便在我的笔记本电脑的声卡（Intel PCH ALC3227 Analog，Ubuntu 18.04）上进行测试，该声卡需要2个通道和16位整数。我还将延迟时间加倍（1秒），关闭重新采样并使演示持续时间更长。这是代码（为了合成而不粘贴运行时错误检查）

#include <alsa/asoundlib.h>
#include <stdlib.h>

static char *device = "hw:1,0"; /* playback device */
snd_output_t *output = NULL;
unsigned char buffer[16*1024]; /* some random data */
int main(void) {
    int err;
    unsigned int i;
    snd_pcm_t *handle;
    snd_pcm_sframes_t frames;
    for (i = 0; i < sizeof(buffer); i++)
        buffer[i] = (unsigned char) (rand() & 0xff);
    snd_pcm_open(&handle, device, SND_PCM_STREAM_PLAYBACK, 0)
    snd_pcm_set_params(handle, SND_PCM_FORMAT_S16_LE,
            SND_PCM_ACCESS_RW_INTERLEAVED, 2, 48000, 0, 1E6);
    // Print actual buffer size
    snd_pcm_hw_params_t *hw_params;
    snd_pcm_hw_params_malloc(&hw_params);
    snd_pcm_hw_params_current(handle, hw_params);
    snd_pcm_uframes_t bufferSize;
    snd_pcm_hw_params_get_buffer_size(hw_params, &bufferSize);
    printf("ALSA buffer size = %li\n", bufferSize);
    // playback
    for (i = 0; i < 256; ++i) {
        frames = snd_pcm_writei(handle, buffer, sizeof(buffer) / 4);
        if (frames < 0)
            frames = snd_pcm_recover(handle, (int) frames, 0);
        if (frames < 0) {
            printf("snd_pcm_writei failed: %s\n", snd_strerror((int) frames));
            break;
        }
        if (frames > 0 && frames < (long) sizeof(buffer) / 4)
            printf("Short write (expected %li, wrote %li)\n",
                    (long) sizeof(buffer) / 4, frames);
    }
    snd_pcm_hw_params_free(hw_params);
    snd_pcm_close(handle);
    return (0);
}

音频工作，但有人可以解释我为什么有时会得到如下输出

ALSA buffer size = 16384
Short write (expected 4096, wrote 9)
Short write (expected 4096, wrote 4080)

表明snd_pcm_writei写的帧少于预期？根据ALSA docs，我知道必须发生信号，但我没有得到原因和信号。

我也尝试将buffer的大小减半，但结果却大致相同。

Answer 1

发生错误时会报告短读取，但某些帧已成功写入。

您应该使用剩余的缓冲区再次调用相同的函数;如果错误不是短暂的，那么它将被报告。 （这个示例代码是错误的;它只是忽略了没有写入缓冲区的剩余部分。）