SystemVerilog 2-bit register decoding problem - system-verilog

Question

I have two 2-bit inputs and an output of 1-bit. So what I'm trying to do is code a next state value with taking AND of the two inputs, then use non-blocking <= to assign that value into register xy_r, which is now two D-flipflops. So I am seeking to get always the XOR-value of the xy_r previous clock edge as my output xor_out. I suppose that the way below is not the right one?
It worked in the simulation but then again in RTL Synthesis I didn't end up with an XOR gate but a third flip-flop so it seems like xor_out is treated as a register.
I suppose I can't use assign outside of the if-else-statements because in that case the output wouldn't follow the xy_r previous state but the present.
Can you please assist me how to solve this issue, if there is a somewhat simple way.
module 2ffsxor
(input logic clk,
input logic rst_n,
input logic [1:0] x_in,
input logic [1:0] y_in,
output logic xor_out
);
logic [1:0] xy;
logic [1:0] xy_r;
always # (posedge clk or negedge rst_n)
begin
if (rst_n == '0)
xy_r <= '0;
else
begin
xy = x_in & y_in;
xy_r <= xy;
xor_out = xy_r[0] ^ xy_r[1];
end
end
endmodule
And here's a schematic what it should be:
Schematic 2ffsxor

Answer 1

Synthesis treats this code as if you had wrote
begin
xy_r <= x_in & y_in;
xor_out <= xy_r[0] ^ xy_r[1];
end
That's two stages of registers. But since you assigned xor_out with a blocking assignment, you have a simulation race condition for any process that tries to read xor_out. The race would be from the result from the previous values of xy_r and the result before that!
If you use an continuous assign statement outside this always block, any process reading xor_out on the clock edge will see the result from the previous state of xy_r.
But why not write this as on register:
always # (posedge clk or negedge rst_n)
begin
if (rst_n == '0)
xor_out <= '0;
else
begin
xy = x_in & y_in;
xor_out <= xy[0] ^ xy[1];
end
end