Skip to main content

is_fully_qualified

pamoja_ros2::name

Function is_fully_qualified 

Source 
pub fn is_fully_qualified(name: &str) -> bool
Expand description

Returns whether a name is fully qualified: valid, absolute, and free of substitutions.

§Arguments

  • name - the candidate name.

§Returns

true if name is valid, starts with /, and contains neither ~ nor {}. Only a fully qualified name can be mapped onto the middleware, because the namespace is already resolved.