CODEWITHSUNDEEP
rustgstreamer
shortcipher3
shortcipher3

Reputation: 1380

Print Element GString Property in Human Readable format

I'm writing a rust application with a simple gstreamer pipeline. I would like to print the stats property of the appsink element in a human readable format.

With this code:

let stats = appsink.get_property("stats").unwrap();
println!("stats: {:?}", stats);

I get:

stats: Value(GString(Foreign(0x7f9c008f00, 101)))

Since that isn't human readable, I tried:

let stats = appsink.get_property("stats").unwrap().get::<GString>();
println!("stats: {:?}", stats);

but got:

stats: Err(GetError { actual: GstStructure, requested: gchararray })

I'm not sure how to interpret the output.

I've looked at this post: gstreamer rust get human readable output for bitrate set on x264enc but it doesn't show how to approach a GString.

Upvotes: 0

Views: 798

Answers (2)

Sebastian Dr&#246;ge
Sebastian Dr&#246;ge

Reputation: 2143

Close to what transistor wrote, the correct way of doing this would be

let stats = appsink.get_property("stats").unwrap();
println!("stats: {:?}", stats.get::<gst::Structure>().expect("not a structure").expect("structure was None"));

You don't have to transform the glib::Value to a String, but you can directly get a gst::Structure from the glib::Value and work on that. Among other things it provides a Debug impl that allows direct printing of it, and various API for accessing the fields, etc. See https://gstreamer.pages.freedesktop.org/gstreamer-rs/gstreamer/structure/struct.StructureRef.htm

Upvotes: 1

transistor
transistor

Reputation: 1660

I was able to sort of reproduce this using the following example:

use gstreamer::prelude::*;

fn main() {
    gstreamer::init().unwrap();
    let source = gstreamer::ElementFactory::make("videotestsrc", Some("source")).expect("Could not create source element.");
    let val = source.get_property("pattern").unwrap();
    println!("{:?}", val);
}

This will attempt to get the pattern property on a generic VideoTestSrc element, and it will print out the string address instead of the actual string. Adding .get::<GString>() to the let val statement will produce a runtime error:

Err(GetError { actual: GstVideoTestSrcPattern, requested: gchararray })

which is telling us that it tried to cast to gchararray but the actual data type of the property is a custom type, GstVideoTestSrcPattern, which is not a string. In your example, the property value has the type GstStructure. It might be possible to use .get::<GstVideoTestSrcPattern>() to get the value of the pattern property and manipulate it as such, but since we want a string here, there's another way using the .transform() method defined on a glib::Value:

let val = source.get_property("pattern").unwrap().transform::<String>().unwrap().get::<String>().unwrap().unwrap();

This is rather unwieldy and it would be advised to do a lot more error checking on the values returned here (for example using the ? operator instead of the .unwrap()s).

The .transform::<String>() call will try to give us a String representation of the property's value, but it gives us a Option<Value> which we must unwrap and convert into an actual String using .get::<String>(), which gives us a Result<Option<String>, GetError> (the inner option is because the string could be NULL). Unwrapping those values gives us a printable string.

There might be a simpler way, but this at least gives you the result. There is more documentation on how to deal with glib Value types here: https://gstreamer.pages.freedesktop.org/gstreamer-rs/glib/value/struct.Value.html But unfortunately it's not very easy to read and doesn't have examples. It might be possible to glean more info from the rust port of the gstreamer tutorials: https://gitlab.freedesktop.org/gstreamer/gstreamer-rs/tree/master/tutorials

Upvotes: 1

Related Questions