Question

我使用脉冲音频的简单C代码进行播放和录制，效果很好。但是当我将它转换为C ++时，它不起作用。我正在粘贴这些代码。请帮忙。 C ++代码不显示任何错误，但不播放任何声音。但C ++代码播放录制的声音。 N.B：我使用的是64位CentOS 6.2

C ++代码：

#ifdef HAVE_CONFIG_H
#include <config.h>
#endif
#include <iostream>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <errno.h>
#include <fcntl.h>
#include <stdint.h>

#include "pulse/simple.h"
#include "pulse/error.h"
using namespace std;
#define BUFSIZE 32
int error;

/* The Sample format to use */


class AudioCapt
{

public: 
    AudioCapt();
    void RecordSound(int argc,char *argv[],pa_simple *s_in,pa_sample_spec &ss,pa_simple *s_out,uint8_t buf[],ssize_t r);
    void PlaybackSound(int argc, char*argv[],pa_simple *s_out,pa_sample_spec &ss,uint8_t buf[],ssize_t r);
};




void AudioCapt::RecordSound(int argc, char*argv[],pa_simple *s_in,pa_sample_spec &ss,pa_simple *s_out,uint8_t buf[],ssize_t r)
{

printf("Audio Capturing \n");
  if (!(s_in = pa_simple_new(NULL, argv[0], PA_STREAM_RECORD, NULL, "record", &ss, NULL, NULL, &error))) {
    fprintf(stderr, __FILE__": pa_simple_new() failed: %s\n", pa_strerror(error));
}

    if (pa_simple_read(s_in, buf, sizeof(buf), &error) < 0) {
        fprintf(stderr, __FILE__": read() failed: %s\n", strerror(errno));
}
printf("Buffer :::: %d\n",buf[0]);

}

void AudioCapt::PlaybackSound(int argc, char*argv[],pa_simple *s_out,pa_sample_spec &ss,uint8_t buf[],ssize_t r)
{
printf("Audio PlayBack \n");
printf("Play Buffer::: %d\n",buf[0]);
if (!(s_out = pa_simple_new(NULL, argv[0], PA_STREAM_PLAYBACK, NULL, "playback", &ss, NULL, NULL,  &error))) {
    fprintf(stderr, __FILE__": pa_simple_new() failed: %s\n", pa_strerror(error));
}

    /* ... and play it (Modified) */
   if (pa_simple_write(s_out, buf, sizeof(buf), &error) < 0) {
       fprintf(stderr, __FILE__": pa_simple_write() failed: %s\n", pa_strerror(error));
}

/* Make sure that every single sample was played */
if (pa_simple_drain(s_out, &error) < 0) {
    fprintf(stderr, __FILE__": pa_simple_drain() failed: %s\n", pa_strerror(error));
}
}

int main(int argc, char * argv[])
{

pa_sample_spec ss;
ss.format = PA_SAMPLE_S16LE;
ss.rate = 44100;
    ss.channels = 2;

pa_simple *s_in, *s_out = NULL;

AudioCapt *m_pMyObject;

for(;;)
{ 
uint8_t buf[BUFSIZE];
    ssize_t r;
m_pMyObject->RecordSound(argc,argv,s_in,ss,s_out,buf,r);
m_pMyObject->PlaybackSound(argc,argv,s_out,ss,buf,r);
}   
return 0;
}

C代码：

#ifdef HAVE_CONFIG_H
#include <config.h>
#endif

#include <stdio.h>
#include <unistd.h>
#include <string.h>
#include <errno.h>
#include <fcntl.h>

#include <pulse/simple.h>
#include <pulse/error.h>

#define BUFSIZE 32

int main(int argc, char*argv[]) {

/* The Sample format to use */
static const pa_sample_spec ss = {
    .format = PA_SAMPLE_S16LE,
    .rate = 44100,
    .channels = 2
};

pa_simple *s_in, *s_out = NULL;
int ret = 1;
int error;


/* Create a new playback stream */
if (!(s_out = pa_simple_new(NULL, argv[0], PA_STREAM_PLAYBACK, NULL, "playback", &ss, NULL, NULL, &error))) {
    fprintf(stderr, __FILE__": pa_simple_new() failed: %s\n", pa_strerror(error));
    goto finish;
}

  if (!(s_in = pa_simple_new(NULL, argv[0], PA_STREAM_RECORD, NULL, "record", &ss, NULL, NULL, &error))) {
    fprintf(stderr, __FILE__": pa_simple_new() failed: %s\n", pa_strerror(error));
    goto finish;
}

for (;;) {
    uint8_t buf[BUFSIZE];
    ssize_t r;

#if 1
    pa_usec_t latency;

    if ((latency = pa_simple_get_latency(s_in, &error)) == (pa_usec_t) -1) {
        fprintf(stderr, __FILE__": pa_simple_get_latency() failed: %s\n", pa_strerror(error));
        goto finish;
    }

    fprintf(stderr, "In:  %0.0f usec    \r\n", (float)latency);

    if ((latency = pa_simple_get_latency(s_out, &error)) == (pa_usec_t) -1) {
        fprintf(stderr, __FILE__": pa_simple_get_latency() failed: %s\n", pa_strerror(error));
        goto finish;
    }

    fprintf(stderr, "Out: %0.0f usec    \r\n", (float)latency);
 #endif

    if (pa_simple_read(s_in, buf, sizeof(buf), &error) < 0) {

        fprintf(stderr, __FILE__": read() failed: %s\n", strerror(errno));
        goto finish;
    }
printf("Buffer :::: %d\n",buf[0]);

    /* ... and play it */
    if (pa_simple_write(s_out, buf, sizeof(buf), &error) < 0) {
        fprintf(stderr, __FILE__": pa_simple_write() failed: %s\n", pa_strerror(error));
        goto finish;
    }
}

/* Make sure that every single sample was played */
if (pa_simple_drain(s_out, &error) < 0) {
    fprintf(stderr, __FILE__": pa_simple_drain() failed: %s\n", pa_strerror(error));
    goto finish;
}

ret = 0;

finish:

if (s_in)
    pa_simple_free(s_in);
if (s_out)
    pa_simple_free(s_out);

return ret;
}

Answer 1

在RecordSound和PlaybackSound中，您使用pa_simple_new初始化临时变量。函数返回后，该值将丢失，并将NULL传递给下一个。

Answer 2

我建议启用编译器检查并修复编译器在此代码上发出的所有错误和警告。

首先，m_pMyObject永远不会被初始化，因此在调用RecordSound时使用它将意味着将未初始化的值传递为＆＃34;这个＆＃34;方法。这通常是件坏事。

在RecordSound和PlaybackSound中，您使用size（buf）告诉库要读/写多少字节。参数buf是指向uint8_t的指针。因此编译器会填充指针的大小（在64位计算机上可能为8）。在这些方法中，您应该使用您拥有的大小参数。在调用中，将sizeof（buf）传递给该参数。

在内存/资源耗尽之前，我不知道库可以创建多少个流。每次调用RecordSound都会创建一个录制流，PlaybackSound会创建一个播放流。这些流永远不会被释放。

总而言之，如果使用未初始化的值来调用RecordSound方法不会导致程序崩溃，它将创建一个记录流并记录几个样本，然后它将创建一个回放流并回放这两个样本。然后它会再次尝试这一切。

为什么用pulseaudio API构造的简单代码不能运行？

2 个答案: