Verilog arithmetic shift implementation.

In summary: In this case, it looks like you need to remove the "result" port from the instantiation, since the module only has 3 ports. This should fix the error and allow the simulation to run properly. In summary, the error is caused by an incorrect number of ports being used in the module instantiation.
  • #1
asd1249jf
Here's my verilog code first of all

module arithmetic_shift(input_num, op, shift_num, result);

input [7:0] input_num;
input [2:0] shift_num;
input op;
output reg [7:0] result;

always @ (*)

case(op)
0 : result = input_num <<< shift_num;
1 : result = input_num >>> shift_num;
default : $display("Error in Arithmetic Shift Module");
endcase
endmodule

module arithmetic_test;
reg [7:0] input_num;
reg op;
reg [2:0] shift_num;
wire [7:0] result;


arithmetic_shift UUT(input_num, op, shift_num, result);

initial
begin
input_num = 0;
op = 0;
shift_num = 1;

#50 input_num = 8'b10101010;
#100 input_num = 8'b11110000;
#150 op = 1;
input_num = 8'b11010001;
shift_num = 3;
end

endmodule

I am using Xilinx to run my simulation.

When I run the syntax check on my code, there are no problems at all. However, when I actually try to run the simulation, I get an error


# Loading work.arithmetic_test
# Loading C:\Modeltech_xe_starter\win32xoem/../xilinx/verilog/xilinxcorelib_ver.arithmetic_shift
# Loading work.glbl
# ** Fatal: (vsim-3365) Testbench.v(12): Too many port connections. Expected 3, found 4.
# Time: 0 ps Iteration: 0 Instance: /arithmetic_test/UUT File: C:/Xilinx/verilog/src/XilinxCoreLib/XFFT1024_V1_1.v
# FATAL ERROR while loading design
# Error loading design



Stating that the instantiation being done on this particular line

arithmetic_shift UUT(input_num, op, shift_num, result);

Does not match the port connections with the actual module. Clearly, this cannot be true as there are 4 ports going in/out of the code.

So here's my question - did I write something wrong on the code to cause this trouble or is Xilinx smoking marijuana?
 
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  • #2
It looks like the issue is that you are instantiating the module with 4 ports, but the module only has 3 ports. You need to make sure that the number of ports you are instantiating the module with matches the number of ports in the module declaration.
 
  • #3



I cannot comment on the specific error you are experiencing with your code. However, I can provide some suggestions to help troubleshoot the issue.

First, I would recommend double-checking the port connections in your instantiation statement to ensure they match the ports in your module. It is possible that there is a typo or mismatch in the port names.

Second, I would suggest checking the syntax of your code to make sure there are no missing or extra semicolons or other syntax errors that could be causing the issue.

Third, you could try running your code on a different simulator to see if the issue persists. This can help determine if the issue is with your code or with the specific simulator you are using.

Lastly, you could also consult with other experts or forums related to Verilog and Xilinx to see if anyone else has encountered a similar issue and how they resolved it.

In conclusion, it is difficult to determine the exact cause of the error without further information or access to your code. However, by carefully checking your code and consulting with other experts, you should be able to identify and resolve the issue.
 

FAQ: Verilog arithmetic shift implementation.

What is Verilog arithmetic shift?

Verilog arithmetic shift is a type of bitwise operation used in hardware design languages, such as Verilog, to shift a binary number to the left or right by a specified number of bits while preserving the sign bit. It is commonly used in digital systems to perform mathematical operations on binary numbers.

How is Verilog arithmetic shift implemented?

Verilog arithmetic shift is typically implemented using the ">>" and "<<" operators. The ">>" operator performs a right shift, while the "<<" operator performs a left shift. The number of bits to shift is specified after the operator. The result is the original number shifted by the specified number of bits.

What is the difference between arithmetic shift and logical shift?

The main difference between arithmetic shift and logical shift is that arithmetic shift preserves the sign bit while shifting, while logical shift does not. In other words, in arithmetic shift, the sign bit is shifted along with the other bits, while in logical shift, the sign bit is replaced with a 0.

What are some common applications of Verilog arithmetic shift?

Verilog arithmetic shift is commonly used in digital systems for various purposes, such as performing multiplication and division by powers of 2, implementing shift registers and counters, and manipulating signed binary numbers.

How can I test my Verilog arithmetic shift implementation?

There are several ways to test a Verilog arithmetic shift implementation. One way is to simulate the code using a Verilog simulator and observe the results. Another way is to use a hardware description language (HDL) testbench to verify the functionality of the implementation. Additionally, it is important to perform boundary and corner case testing to ensure the implementation works correctly in all scenarios.

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