verilog – Capturing the right posedge clock in Quartus waveform

I am using Quartus Prime Lite 19.1.0.

module memory_address_register1 #(
    parameter                       ADDR_WIDTH = 4
    input                           clk, rst, load,
    input       [ADDR_WIDTH-1:0]    add_in,
    output reg     [ADDR_WIDTH-1:0]    add_out

    always @(posedge clk) begin
        if (rst) begin
            add_out <= 4'b0000;
        end else if (load) begin
            add_out <= add_in;
        end else begin
            add_out <= add_out;

module mmr_tb();
    reg clk, rst, load;
    reg [3:0] add_in;
    wire [3:0] add_out;
    initial begin
        clk <= 1'b0;
        rst <= 1'b0;
        load <= 1'b0;
    memory_address_register1 mmr (.clk(clk), .rst(rst), .load(load), .add_in(add_in), .add_out(add_out));
    always #10 clk = ~clk;
    initial begin
        #20 add_in <= 4'd2;
        #10 add_in <= 4'd3;
        load <= 1'b1;
        #30 add_in <= 4'd6;
        #10 load <= 1'b1;


It the output (add_out) accurate? Should the output (add_out) at t=70.0ns be “6” or “7”?

  1. If the expected output is “6”, can anyone explain why is that?

img1: posedge clk output value from previous clk cycle

img2: posedge clk output value from previous clk cycle

I ran the testbench using modelsim, and I am able to get the expected output I wanted (output on the exact clock edge), but is it expected?

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