Analyzing unexpected output from Verilog bit unpacking and reassembly logic

Question

I'm working on a Verilog task that rearranges bits from a 312-bit word into a new 312-bit format using 8-bit temporary storage (temp[39]). Below is a simplified version of my code:

`define NUM_OF_BANK_LINES (1024*3)
`define RAM_MODEL_DEPTH 1024
module test;
 reg i_CLK = 0;
 reg [311:0] RAM_BANK_LO[`RAM_MODEL_DEPTH * 3];
 reg [311:0] MY_RAM_BANK_LO[`RAM_MODEL_DEPTH * 3];
 int i, j;
 always #(1.25) i_CLK = ~i_CLK;
 task mem_task(
 input [311:0] IN_MEM_BANK [`NUM_OF_BANK_LINES],
 output reg [311:0] OUT_MEM_BANK [`NUM_OF_BANK_LINES]
 );
 reg [7:0] temp[39];
 for (int row = 0; row < `NUM_OF_BANK_LINES; row++) begin
 for (int loc = 0; loc < 39; loc++) begin
 temp[loc][0] = IN_MEM_BANK[row][39*0 + loc];
 temp[loc][1] = IN_MEM_BANK[row][39*1 + loc];
 temp[loc][2] = IN_MEM_BANK[row][39*2 + loc];
 temp[loc][3] = IN_MEM_BANK[row][39*3 + loc];
 temp[loc][4] = IN_MEM_BANK[row][39*4 + loc];
 temp[loc][5] = IN_MEM_BANK[row][39*5 + loc];
 temp[loc][6] = IN_MEM_BANK[row][39*6 + loc];
 temp[loc][7] = IN_MEM_BANK[row][39*7 + loc];
 end
 OUT_MEM_BANK[row] = {
 temp[38], temp[37], temp[36], temp[35], temp[34], temp[33],
 temp[32], temp[31], temp[30], temp[29], temp[28], temp[27],
 temp[26], temp[25], temp[24], temp[23], temp[22], temp[21],
 temp[20], temp[19], temp[18], temp[17], temp[16], temp[15],
 temp[14], temp[13], temp[12], temp[11], temp[10], temp[9],
 temp[8], temp[7], temp[6], temp[5], temp[4], temp[3],
 temp[2], temp[1], temp[0]
 };
 for (j = 0; j < 39; j++)
 $display("temp[%0d] = %0h", j, temp[j]);
 $display("OUT_MEM_BANK[%0d] = %0h", row, OUT_MEM_BANK[row]);
 end
 endtask
 initial begin
 for (i = 0; i < 1000; i++) begin
 RAM_BANK_LO[i] = i;
 end
 mem_task(RAM_BANK_LO, MY_RAM_BANK_LO);
 end
endmodule

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The problem:

When temp[0] = 1 and temp[1] = 1, and all other temp[2] to temp[38] are zero, I expect:

OUT_MEM_BANK[3] = 8'b00000011 = 0x03

However, what I actually get is:

OUT_MEM_BANK[3] = 0x65 ('h101)

enter image description here

Question:

Why am I seeing OUT_MEM_BANK[3] = 101 instead of 0x03, even though only temp[0] and temp[1] are 1?

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For helping to understand my problem I attached a link: https://www.edaplayground.com/x/EFth

Answer 1

You are seeing the correct simulation output:

OUT_MEM_BANK[3] = 101

You declared OUT_MEM_BANK as 312 bits wide, and you declared temp as 8 bits wide.

Array elements 2-38 of temp are all 0, which is why all the MSBs of OUT_MEM_BANK[3] are 0. You understand this part.

You set the temp[0] byte to 1, which is the same as 0x01, or in Verilog syntax 8'h01.

You set temp[1] to 1 also.

When you concatenate {temp[1], temp[0]}, it is the same as {8'h01, 8'h01}. This simplifies to 16'h0101. Since you $display using the format %0h, the 0 instructs the simulator to print the hex values without any leading 0's. This is why it prints the hex value as 101.

Change your code like this to see all 312 bits:

 $display("OUT_MEM_BANK[%0d] = %h", row, OUT_MEM_BANK[row]);

Each hex digit represents 4 bits. Two hex digits are needed to represent 8 bits.

As a demonstration, try setting temp[0] to a larger decimal value, like 255. Do the same for temp[1].

Here is an example on EDA Playground.

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0x65 is not the same as 'h101. 'h101 is the same as 0x101.