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[求助] 同样的代码 VCA和questasim仿真得到结果不一样

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发表于 2022-4-3 13:46:51 | 显示全部楼层 |阅读模式

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本帖最后由 不知名的wei 于 2022-4-3 13:56 编辑

在图中勾出来了
在questasim中 才是我想要的波形。

questasim翻转波形

questasim翻转波形

VCS仿真波形

VCS仿真波形
发表于 2022-4-5 02:34:18 | 显示全部楼层
Share the code please
发表于 2022-4-5 08:04:04 | 显示全部楼层
貌似有一种原因是赋值上的区别,理论上两种工具都应在边沿(如上升沿)之后赋值,但是在具体仿真时两个工具实现有点差别(比如一个在上升沿前?),因此最好加上如#1来消除这种不确定性
 楼主| 发表于 2022-4-5 13:48:02 | 显示全部楼层


ashish 发表于 2022-4-5 02:34
Share the code please


module check
(
input en,
input clk,
input rst,
input [15:0]datain,
output reg [4:0] out
);



function [2:0] checknum;
input [3:0] a;
begin
   if(a[3]==1'b1)
      checknum=3'd0;
else  if(a[2]==1'b1)
       checknum=3'd1;
else  if(a[1]==1'b1)
       checknum=3'd2;
else  if(a[0]==1'b1)
       checknum=3'd3;
  else
       checknum=3'd4;
end
endfunction


wire [2:0] num3,num2,num1,num0;
reg  [4:0] out_r;


assign  num3=checknum(datain[15:12]);
assign  num2=checknum(datain[11:8]);
assign  num1=checknum(datain[7:4]);
assign  num0=checknum(datain[3:0]);


//out_r是组合逻辑输出 所以和时钟边沿无关
always@(num3,num2,num1,num0)
begin
    if(num3<3'd4)
        out_r<= {2'd0,num3};
    else  if(num2<3'd4)
       out_r<= 5'd4+{2'd0,num2};
    else if(num1<3'd4)
       out_r<= 5'd8+{2'd0,num1};
    else
       out_r<= 5'd12+{2'd0,num0};       
end



always@(posedge clk or negedge rst)
begin
   if(rst==1'b0)
   out<= 5'd0;
   else if(en==1'b1)
   out<= out_r;
end

endmodule
//这是RTL代码

 楼主| 发表于 2022-4-5 13:52:59 | 显示全部楼层


qleenju 发表于 2022-4-5 08:04
貌似有一种原因是赋值上的区别,理论上两种工具都应在边沿(如上升沿)之后赋值,但是在具体仿真时两个工具 ...


加过  没啥用加不加在 questasim中完全正确无误的 在vcs中就是不对。
大佬 我的代码在2楼可查看
谢谢

发表于 2022-4-6 01:32:10 | 显示全部楼层
Always use blocking assignments for combo logic. Try this code and let un know.




  1. module check(
  2.              input            en,
  3.              input            clk,
  4.              input            rst,
  5.              input [15:0]     datain,
  6.              output reg [4:0] out
  7.              );

  8.    function [2:0] checknum;
  9.       input [3:0] a;
  10.       begin
  11.          if(a[3]==1'b1)
  12.            checknum = 3'd0;
  13.          else if(a[2]==1'b1)
  14.            checknum = 3'd1;
  15.          else if(a[1]==1'b1)
  16.            checknum = 3'd2;
  17.          else if(a[0]==1'b1)
  18.            checknum = 3'd3;
  19.          else
  20.            checknum = 3'd4;
  21.       end
  22.    endfunction


  23.    wire [2:0] num3,num2,num1,num0;
  24.    reg [4:0 ] out_r;


  25.    assign num3 = checknum(datain[15:12]);
  26.    assign num2 = checknum(datain[11:8]);
  27.    assign num1 = checknum(datain[7:4]);
  28.    assign num0 = checknum(datain[3:0]);


  29. //out_r is a combinational logic output so it has nothing to do with the clock edge
  30.    always@(num3,num2,num1,num0) begin
  31.       if(num3<3'd4)
  32.         out_r = {2'd0,num3};
  33.       else if(num2<3'd4)
  34.         out_r = 5'd4+{2'd0,num2};
  35.       else if(num1<3'd4 )
  36.         out_r = 5'd8+{2'd0,num1};
  37.       else
  38.         out_r = 5'd12+{2'd0,num0};      
  39.    end


  40.    always@(posedge clk or negedge rst) begin
  41.       if (rst==1'b0)
  42.         out <= 5'd0;
  43.       else if (en==1'b1)
  44.         out <= out_r;
  45.    end
  46. endmodule // check
  47. //this is the RTL code



复制代码
 楼主| 发表于 2022-4-6 09:36:35 | 显示全部楼层


ashish 发表于 2022-4-6 01:32
Always use blocking assignments for combo logic. Try this code and let un know.


很抱歉  使用您给的代码以后  questasim中是正确的  但是VCS还是不对。
我不知道是什么原因
我把包含测试平台的代码发给您  

谢谢您

代码.rar

7.56 KB, 下载次数: 1 , 下载积分: 资产 -2 信元, 下载支出 2 信元

发表于 2022-4-7 02:27:52 | 显示全部楼层
Fixed driver.sv for vcs now vcs and questasim results will match

code_fixed.rar

129.26 KB, 下载次数: 1 , 下载积分: 资产 -2 信元, 下载支出 2 信元

code_fixed

 楼主| 发表于 2022-4-7 09:25:00 | 显示全部楼层
本帖最后由 不知名的wei 于 2022-4-7 09:52 编辑


ashish 发表于 2022-4-7 02:27
Fixed driver.sv for vcs now vcs and questasim results will match


是的 正确了  在driver.sv中给激励的时候 需要非阻塞赋值         然而 在quetasim中    采用阻塞和非阻塞结果都是一样的
这是为什么了?


发表于 2022-4-7 14:29:34 | 显示全部楼层
This is because of the way the simulators are implemented. The way vcs and questa handles blocking and non blocking assignments is not the same. Here is introduction how internally the simulators work..
This chapter is from the book
Hardware Design Verification: Simulation and Formal Method-Based ApproachesHardware Design Verification: Simulation and Formal Method-Based Approaches
1.6 A Quick Overview of Verilog Scheduling and Execution Semantics
On the assumption that you have a good understanding of Verilog, we will not review the general syntax of Verilog. However, we will review the more advanced features of the language—namely, scheduling and execution semantics—they appear in various places throughout this book. If you are well versed in Verilog, you can skip this section.

1.6.1 Concurrent Processes
Verilog is a parallel hardware descriptive language. It differs from a software language such as C/C++ in that it models concurrency of hardware. Components in a circuit always execute in parallel, whereas a software program executes one statement at time—in a serial fashion. To model hardware concurrency, Verilog uses two types of data structures: continuous assignment and procedural block. A continuous assignment executes whenever a variable in the right side of the assignment changes. As the name implies, a continuous assignment continuously watches changes of variables on the right side and updates the left-side variables whenever changes occur. This behavior is consistent with that of a gate, which propagates inputs to outputs whenever inputs change. An example of a continuous assignment is shown here. The value of v is updated whenever the value of x, y, or z changes:

assign v = x + y + z;
A procedural block consists of statements called procedural statements. Procedural statements are executed sequentially, like statements in a software program, when the block is triggered. A procedure block has a sensitivity list containing a list of signals, along with the signals' change polarity. If a signal on a sensitivity changes according to its prescribed polarity, the procedural block is triggered. On the other hand, if a signal changes that is not on the sensitivity list but is present in the right-hand side of some procedural statements of the block, the procedural statements will not be updated. An example of a procedural block is shown here:

always @(posedge c or d) begin
   v = c + d + e;
   w = m - n;
end
The signals on the sensitivity list are c and d. The polarity for c is positive edge, and no polarity is specified for d, which by default means either positive or negative edge. This block is executed whenever a positive event occurs at c. A positive edge is one of the following: a change from 0 to 1, X, or Z; or from X or Z to 1. The block is also triggered when an event occurs at d, positive or negative edge. A negative edge is one of the following: a change from 1 to either 0, X, or Z; or from X or Z to 0. Changes on variables e, m, or n will not cause the block to execute because these three variables are not on the sensitivity list.

In a way, a procedural block is a generalization of a continuous assignment: Any variable change on the right side of a continuous assignment triggers evaluation, whereas in a procedural block only changes in the variables on the sensitivity list can trigger evaluation, and the changes must conform to the polarity specification. Any continuous assignment can be rewritten as a procedural block, but not vice versa. The following procedural block and continuous assignment are equivalent:

// continuous assignment
assign v = x + y + z;

// equivalent procedural block
always @(x or y or z)
   v = x + y + z;
However, there is no equivalent continuous assignment for the following procedural block, because rising or falling changes are not distinguished in continuous assignment:

always @(posedge x)
   y = x;
1.6.2 Nondeterminism
When multiple processes execute simultaneously, a natural question to ask is how these processes are scheduled. When multiple processes are triggered at the same time, the order in which the processes are executed is not specified by Institute of Electrical and Electronics Engineers (IEEE) standards. Thus, it is arbitrary and it varies from simulator to simulator. This phenomenon is called nondeterminism. Let's look at two common cases of nondeterminism that are caused by the same root cause but that manifest in different ways.

The first case arises when multiple statements execute in zero time, and there are several legitimate orders in which these statements can execute. Furthermore, the order in which they execute affects the results. Therefore, a different order of execution gives different but correct results. Execution in zero time means that the statements are evaluated without advancing simulation time. Consider the following statements:

always @(d)
   q = d;

assign q = ~d;
These two processes, one procedural block and one continuous assignment, are scheduled to execute at the same time when variable d changes. Because the order in which they execute is not specified by the IEEE standard, two possible orders are possible. If the always block is evaluated first, variable q is assigned the new value of d by the always block. Then the continuous assignment is executed. It assigns the complement of the new value of d to variable q. If the continuous assignment is evaluated first, q gets the complement of the new value of d. Then the procedural assignment assigns the new value (uncomplemented) to q. Therefore, the two orders produce opposite results. This is a manifestation of nondeterminism.

Another well-known example of nondeterminism of the first type occurs often in RTL code. The following D-flip-flop (DFF) module uses a blocking assignment and causes nondeterminism when the DFFs are instantiated in a chain:

module DFF (clk, q, d);
input clk,  d;
output q;
reg q;

always @(posedge clk)
   q = d; // source of the problem

endmodule

module DFF_chain;

DFF dff1(clk, q1, d1); // DFF1
DFF dff2(clk, q2, q1); // DFF2

endmodule
When the positive edge of clk arrives, either DFF instance can be evaluated first. If dff1 executes first, q1 is updated with the value of d1. Then dff2 executes and makes q2 the value of q1. Therefore, in this order, q2 gets the latest value of d1. On the other hand, if dff2 is evaluated first, q2 gets the value of q1 before q1 is updated with the value of d1. Either order of execution is correct, and thus both values of q2 are correct. Therefore, q2 may differ from simulator to simulator.

A remedy to this race problem is to change the blocking assignment to a nonblocking assignment. Scheduling of nonblocking assignments samples the values of the variables on the right-hand side at the moment the nonblocking assignment is encountered, and assigns the result to the left-side variable at the end of the current simulation time (for example, after all blocking assignments are evaluated). A more in-depth discussion of scheduling order is presented in the section "Scheduling Semantics." If we change the blocking assignment to a nonblocking assignment, the output of dff1 always gets the updated value of d1, whereas that of dff2 always gets the preupdated value of q1, regardless of the order in which they are evaluated.

Nonblocking assignment does not eliminate all nondeterminism or race problems. The following simple code contains a race problem even though both assignments are nonblocking. The reason is that both assignments to variable x occur at the end of the current simulation time and the order that x gets assigned is unspecified in IEEE standards. The two orders produce different values of x and both values of x are correct:

always @(posedge clk)
begin
   x <= 1'b0;
   x <= 1'b1;
end
The second case of nondeterminism arises from interleaving procedural statements in blocks executed at the same time. When two procedural blocks are scheduled at the same time, there is no guarantee that all statements in a block finish before the statements in the other block begin. In fact, the statements from the two blocks can execute in any interleaving order. Consider the following two blocks:

always @(posedge clk) // always block1
begin
   x = 1'b0;
   y = x;
end

always @(posedge clk) // always block2
begin
   x = 1'b1;
end
Both always blocks are triggered when a positive edge of clk arrives. One interleaving order is

x = 1'b0;
y = x;
x = 1'b1;
In this case, y gets 0.

Another interleaving order is

x = 1'b0;
x = 1'b1;
y = x;
In this case, y gets 1.

There is no simple fix for this nondeterminism. The designer has to reexamine the functionality the code is supposed to implement and, more often than not, nondeterminism is not inherent in the functionality but is introduced during the implementation process.

1.6.3 Scheduling Semantics
Having discussed nondeterminism, it is natural for us to examine the scheduling semantics in Verilog (that is, the order in which events are scheduled to execute). Events at simulation time are stratified into five layers of events in the order of processing:

Active
Inactive
Nonblocking assign update
Monitor
Future events
Active events at the same simulation time are processed in an arbitrary order. An example of an active event is a blocking assignment. The processing of all the active events is called a simulation cycle.

Inactive events are processed only after all active events have been processed. An example of an inactive event is an explicit zero-delay assignment, such as #0 x = y, which occurs at the current simulation time but is processed after all active events at the current simulation time have been processed.

A nonblocking assignment executes in two steps. First, it samples the values of the right-side variables. Then it updates the values to the left-side variables. The sampling step of a nonblocking assignment is an active event and thus is executed at the moment the nonblocking statement is encountered. A nonblocking assign update event is the updating step of a nonblocking assignment and is executed only after both active and inactive events at the current simulation time have been processed.

Monitor events are the execution of system tasks $monitor and $strobe, which are executed as the last events at the current simulation time to capture steady values of variables at the current simulation time. Finally, events that are to occur in the future are the future events.
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