C++ gstreamer-0.10打印管道字符串

C++ gstreamer-0.10打印管道字符串,c++,gstreamer,gstreamer-0.10,C++,Gstreamer,Gstreamer 0.10,我想打印出gstreamer-0.10元素的管道。如何做到这一点 外部开发人员为我们编写的代码: static GstElement* jpgPipeline = NULL; pipedef = g_strdup_printf("appsrc name=jpgsrc ! ffmpegcolorspace %s %s %s %s %s ! ffmpegcolorspace ! jpegenc quality=%i ! multifilesink name=jpgsink location=\"

我想打印出gstreamer-0.10元素的管道。如何做到这一点

外部开发人员为我们编写的代码:

static GstElement* jpgPipeline  = NULL;
pipedef = g_strdup_printf("appsrc name=jpgsrc ! ffmpegcolorspace %s %s %s %s %s  ! ffmpegcolorspace ! jpegenc quality=%i ! multifilesink name=jpgsink location=\"image.jpg\"",cropStr, scaleStr, sharpenStr,vbStr,gammaStr,quality);
jpgPipeline = gst_parse_launch (pipedef, &error);
bus = gst_pipeline_get_bus(GST_PIPELINE(jpgPipeline));
gst_element_set_state (jpgPipeline, GST_STATE_PLAYING);
我试图通过以下方式查看管道:

g_print(gst_element_info(jpgPipeline));
但当我试图编译它时,会收到很多警告

main.c:331:2: warning: implicit declaration of function ‘gst_element_info’ [-Wimplicit-function-declaration]
  g_print(gst_element_info(jpgPipeline));
  ^
main.c:331:2: warning: passing argument 1 of ‘g_print’ makes pointer from integer without a cast [enabled by default]
In file included from /usr/include/glib-2.0/glib.h:62:0,
                 from /usr/include/gstreamer-0.10/gst/gst.h:27,
                 from main.c:5:
/usr/include/glib-2.0/glib/gmessages.h:265:17: note: expected ‘const gchar *’ but argument is of type ‘int’
 void            g_print                 (const gchar    *format,
                 ^
main.c:331:2: warning: format not a string literal and no format arguments [-Wformat-security]
  g_print(gst_element_info(jpgPipeline));
main.c:(.text.startup+0x651): undefined reference to `gst_element_info'

第一点也是最重要的一点:请使用1.0,0.10已经过时多年了

gst\u元素\u信息
可能不存在。关于管道,您希望以字符串形式打印什么内容?其中的元素是什么?您希望如何表示链接


也许使用点表示会更好。在这里找到一些说明:或者

pipedef已经是您的管道,只需g_print(pipedef)即可。