[PATCH v6 1/9] of: property: add of_graph_get_next_port()

Rob Herring robh at kernel.org
Tue Oct 8 14:18:34 UTC 2024


On Thu, Sep 26, 2024 at 12:00:12AM +0000, Kuninori Morimoto wrote:
> We have endpoint base functions
> 	- of_graph_get_next_device_endpoint()
> 	- of_graph_get_device_endpoint_count()
> 	- for_each_of_graph_device_endpoint()
> 
> Here, for_each_of_graph_device_endpoint() loop finds each endpoints
> 
> 	ports {
> 		port at 0 {
> (1)			endpoint {...};
> 		};
> 		port at 1 {
> (2)			endpoint {...};
> 		};
> 		...
> 	};
> 
> In above case, it finds endpoint as (1) -> (2) -> ...
> 
> Basically, user/driver knows which port is used for what, but not in
> all cases. For example on flexible/generic driver case, how many ports
> are used is not fixed.
> 
> For example Sound Generic Card driver which is used from many venders
> can't know how many ports are used. Because the driver is very
> flexible/generic, it is impossible to know how many ports are used,
> it depends on each vender SoC and/or its used board.
> 
> And more, the port can have multi endpoints. For example Generic Sound
> Card case, it supports many type of connection between CPU / Codec, and
> some of them uses multi endpoint in one port.
> Then, Generic Sound Card want to handle each connection via "port"
> instead of "endpoint".
> But, it is very difficult to handle each "port" via existing
> for_each_of_graph_device_endpoint(). Getting "port" via of_get_parent()
> from "endpoint" doesn't work. see below.
> 
> 	ports {
> 		port at 0 {
> (1)			endpoint at 0 {...};
> (2)			endpoint at 1 {...};
> 		};
> 		port at 1 {
> (3)			endpoint {...};
> 		};
> 		...
> 	};
> 
> In other case, we want to handle "ports" same as "port" for some reasons.
> 
> 	node {
> =>		ports at 0 {
> 			port at 0 { ... };
> 			port at 1 { ... };
> 			...
> 		};
> =>		ports at 1 {

As I said before. No where is this documented. We're not going to add 
common helpers for something undocumented and non-standard. Plus, this 
patch is doing a lot more than $subject says.

Rob


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