Difficulty: Hard · Puzzle 4 of 6 · Topic: Factory Override Puzzles
An exact instance override and a wildcard instance override both match one path. Which type is created?
class base_item extends uvm_object;
`uvm_object_utils(base_item)
function new(string name="base_item"); super.new(name); endfunction
endclass
class dbg_item extends base_item;
`uvm_object_utils(dbg_item)
function new(string name="dbg_item"); super.new(name); endfunction
endclass
class perf_item extends base_item;
`uvm_object_utils(perf_item)
function new(string name="perf_item"); super.new(name); endfunction
endclass
initial begin
base_item::type_id::set_inst_override(
perf_item::get_type(),
"uvm_test_top.env.agent*.seqr.main_phase"
);
base_item::type_id::set_inst_override(
dbg_item::get_type(),
"uvm_test_top.env.agent0.seqr.main_phase"
);
$display("a0=%s",
base_item::type_id::create("a0", null, "uvm_test_top.env.agent0.seqr.main_phase").get_type_name());
$display("a1=%s",
base_item::type_id::create("a1", null, "uvm_test_top.env.agent1.seqr.main_phase").get_type_name());
endExact path match is more specific than wildcard match.
1) agent0 path matches both wildcard and exact entries.
2) Factory selects the more specific exact instance override for agent0 -> `dbg_item`.
3) agent1 only matches wildcard -> `perf_item`.Answer: `a0=dbg_item` and `a1=perf_item` due to exact instance-path precedence over wildcard.
Assuming wildcard registration order alone determines result leads to incorrect predictions.
How would you debug this at runtime using factory print/debug APIs?