Ошибка 10170 quartus

Hi, i got some trouble for the code i developed: 

when i execute this code: 

if(rst==1’b1) 

begin 

38. cs [0] = 4’b0; 

39. cs [1] = 4’b0; 

40. cs [2] = 4’b0; 

41. cs [3] = 4’b0; 

42. cs [4] = 4’b0; 

43. s [5] = 4’b0; 

end 

 

then compile and i got that syntax: 

Error (10170): Verilog HDL syntax error at digitalclock.v(39) near text «=»; expecting «.», or an identifier 

Error (10170): Verilog HDL syntax error at digitalclock.v(40) near text «=»; expecting «.», or an identifier 

Error (10170): Verilog HDL syntax error at digitalclock.v(41) near text «=»; expecting «.», or an identifier 

Error (10170): Verilog HDL syntax error at digitalclock.v(42) near text «=»; expecting «.», or an identifier 

Error (10170): Verilog HDL syntax error at digitalclock.v(43) near text «=»; expecting «.», or an identifier 

please note line 38! 

please help me solved this error, thank you! :)

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    You may encounter an error indicating syntax error 10170 verilog hdl. It turns out that there are several ways to solve this problem, and this is what we will now look at. / Error (10170): Verilog HDL syntax error in .v (line_number) next to text message “,”; expects an operand. Due to your issue with Quartus® II software type 13.1 and above, you may receive the following error when compiling a Verilog HDL file converted from a block design file (.bdf).

    Due to the situation in Quartus® II software version 13.1 and shortly thereafter, you may receive the following error message when Verilog compiles an HDL file converted from an HDL file: block construction (.bdf).

    The reason for the error is that the entire generated Verilog HDL file contains almost all the extra commas in mov connections.

    The reason for your syntax error can be described in such a way that you cannot simply write:

      Product [7: 4] matches 4'b0000; 
      assign product [7: 4] is 4'b0000; 

    But if you are not using System Verilog (and your old-fashioned coding style might suggest that you are not), you will find that

      product assignment [7: 4] includes 4'b0000; 

    also doesn’t compile, because the sad victim of the assign statement should automatically be wire , not reg . And if you replace product with a new stream , you will find yourself getting instructions and errors like this:

      product means product >> 1; // move well and set the high bit to 0Product [7: 3] = product denion [7: 3] + multiplicable [4: 0]; // add 5 bits, so we'll probably deal with carry 
      product = fabric >> 1; // move all the way to the right 

    error 10170 verilog hdl syntax error

    because you cannot assign a full stream in a always (or initial ) block.

    You are starting to design a “shift and add art” multiplier and you probably really want to initialize the product at the beginning of the calculation. (Assuming you are creating syntax) lines

      (assign) [7: 4] item = 4'b0000;(Assignment) the product [3: 0] is equal to the multiplier [3: 0]; 

    run product constantly, every time; they do not initialize product . This is where you design hardware, not software.

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  •  1234th56th7th8thnineteneleven12th13th14thfifteen1617th18th19th20th21 years22nd232425262728 year2930th31 year32333435 year363738394041 years4243 years old44 years45464748495051525354555657 years old5859

    error 10170 verilog hdl syntax error

     module kj (j1, l1, j, k, clock, reset, q, qb, q1, qb1, b);Input j1, l1, j, k, clock, reset;Output reg q1, qb1;Reg exit. [3: 0] q, qb, b, a;always @ (negative hours)beginCase (reset, j1, l1)3'b100: q1 = q1;3'b101: q1 = 0;3'b110: q1 = 1;3'b111: q1 = ~ q1;Default : q1 = 0;Back coverqb1 <= ~ q1;[email protected] *beginif (q1 == q1)beginkl JK1 (j, k, hours, reset, q [0], qb [0]);kl JK2 (j, k, q [0], reset, q [1], qb [1]);kl JK3 (j, k, q [1], reset, q [2], qb [2]);kl JK4 (j, k, q [2], reset, q [3], qb [3]);endendotherwise, if (q1 == 0)beginkl JK5 (j, k, d, reset, q [0], qb [0]);kl JK6 (j, k, q [0], reset, q [1], qb [1]);kl JK7 (j, k, q [1], reset, q [2], qb [2]);kl JK8 (j, k, q [2], reset, q [3], qb [3]);endAotherwise, if (q1 == 1)beginalways @ (reset)beginif (reset)q <= 4'b0000;otherwise, if (q <4'b0101)d <= d + 1;anotherb = q [1] && q [3];endendAotherwise, if (q1 == ~ q1)beginalways @ (clock setting)beginif (reset)q <= 4'b0000;otherwise, if (q <4'b0011)qQ + 1;AnotherA = q [2] & q [3];endendFinal module 

    <=

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    Error 10170 Compilation format error

    I am the latest in Verilog. System error of the if statement.
    Can a person help me by rejecting my mistake?

    This is partially related code I wrote. The following
    error occurs

    Error (10170): Verilog HDL syntax error in seqdet.v (24) next to if content;
    Waiting for an identifier («if» is a reserved keyword), quantity, system or
    backchi, or «(» and also «{» or unary operator,

    current_state is in register type, and reset_state was initialized to 3’b000 using the
    parameter instruction.

    Post by Jughead
    I’m new to Verilog. Collect error for if statement.
    Can the player help me by pointing out my mistake ?. [2: 0]
    reg next_state, current_state;
    the reset_state parameter means 3’b000;
    case (data)
    {
    if (current_state == reset_state)
    begin

    next_state means reset_state;

    end
    This is part because of the code I wrote. There is the following error

    Verilog error (10170): HDL syntax error in seqdet.v (24) next to the anchor text «if»;
    Waiting for an identifier («if» is a keyword of any type of reserved identifier), or #, or
    of a system task, or «(«, with «{«, or a unary operator,
    current_state is associated with a type register, and reset_state is initialized to 3 ‘b000 with parameter declaration. Out
    thanks,
    aravind

    This case statement did not have large opening and closing
    statements, and it was enclosed in curly braces, which, in my opinion, was unnecessary. Not sure if I’m right. But now I am notI understand all our mistakes.

    used for bitstring. Use start-end instead.
    Don’t forget the «endcase».

    used for both bitstring. Use start-end instead. Film Not such an «extreme case».

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    Fehler 10170 Verilog Hdl Syntaxfehler
    Fout 10170 Verilog Hdl Syntaxisfout
    오류 10170 Verilog Hdl 구문 오류
    Erro 10170 Erro De Sintaxe Hdl Verilog
    Erreur 10170 Erreur De Syntaxe Verilog Hdl
    Fel 10170 Verilog Hdl Syntaxfel
    Errore 10170 Errore Di Sintassi Verilog Hdl
    Blad 10170 Verilog Blad Skladni Hdl
    Oshibka 10170 Sintaksicheskaya Oshibka Verilog Hdl
    Error 10170 Error De Sintaxis De Verilog Hdl

    Пример-копипаста из книги:

    module test2(input logic [3:0] a, input logic en, output tri [3:0] y);
    assign y = en ? a : 4’bz;
    endmodule

    Ошибки:

    Error (10170): Verilog HDL syntax error at test2.v(2) near text
    Error (10170): Verilog HDL syntax error at test2.v(2) near text «»; expecting «;»
    Error (10112): Ignored design unit «test2» at test2.v(1) due to previous errors

    WTF?


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    По умолчанию во многих CAE синтез верилога идёт подразумевая настройку синтаксиса verilog-95,
    вам следует изменить настройки (по описанию портов видно, что это минимум verilog-2001, а по резервированному слову logic — SV)


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    Минуточку внимания

    Codes:
    module adder(
    input [31:0] operand1,
    input [31:0] operand2,
    input cin,
    output [31:0] result,
    output cout
    );
    assign {cout,result} = operand1 + operand2 + cout;
    endmodule

    Error Messages:

    Error (10170): Verilog HDL syntax error at Verilog1.v(8) near text: “cout”; expecting “highz0”, or “highz1”, or “large”, or “medium”, or “pull0”, or “pull1”, or “small”, or “strong0”, or “strong1”, or “supply0”, or “weak0”, or “weak1”. Check for and fix any syntax errors that appear immediately before or at the specified keyword. The Intel FPGA Knowledge Database contains many articles with specific details on how to resolve this error. Visit the Knowledge Database at https://www.altera.com/support/support-resources/knowledge-base/search.html and search for this specific error message number.
    Error (10759): Verilog HDL error at Verilog1.v(8): object result declared in a list of port declarations cannot be redeclared within the module body
    Error (10112): Ignored design unit “adder” at Verilog1.v(1) due to previous errors
    Error: Quartus Prime Analysis & Synthesis was unsuccessful. 3 errors, 1 warning
    Error: Peak virtual memory: 4702 megabytes
    Error: Processing ended: Sat Nov 27 15:57:22 2021
    Error: Elapsed time: 00:00:12
    Error: Total CPU time (on all processors): 00:00:27
    Error (293001): Quartus Prime Full Compilation was unsuccessful. 5 errors, 1 warning


    Solution:

    Select the option in the red box.

    Done!

    Read More:

    I am trying to read a .mif file and this is the error I am getting —

    Error (10170): Verilog HDL syntax error at sine3_test1.MIF near
    text: p
    . Check for and fix any syntax errors that appear immediately
    before or at the specified keyword. The Intel FPGA Knowledge Database
    contains many articles with specific details on how to resolve this
    error. Visit the Knowledge Database at
    https://www.altera.com/support/support-resources/knowledge-base/search.html
    and search for this specific error message number.

    The error seems to be in first line.
    My MIF file has 8-bit data width and 256 words depth.

    memory address = UNS (unisgned decimal)
    memory data = hexadecimal
    

    enter image description here

    My source code is the following:

    reg [7:0]memory[0:255];
    begin
        $readmemh("sine3_test1.MIF",memory);
    end
    

    I removed white spaces and added a few missing memory block numbers.
    Currently, I have it stored in my project folder (although not part of the Quartus project).

    toolic's user avatar

    toolic

    57.9k17 gold badges75 silver badges117 bronze badges

    asked Apr 18 at 5:47

    Maverick's user avatar

    The format of the input file is incorrect for $readmemh. The $readmemh system task expects a file to just have numbers in hexadecimal format, not words like DEPTH, or address/data pairs like 0:78;.

    You can not directly read the mif file using $readmemh. You need to create another file with just the data in it, like sine3_test1.hex:

    78
    7a
    7d
    

    Then use:

    $readmemh("sine3_test1.hex", memory);
    

    For the complete specification of the input file format, refer to IEEE Std 1800-2017, section 21.4 Loading memory array data from a file. The format does support address information, but it does not seem necessary in this case.


    You could read the mif file using $fopen (and related functions) instead of $readmemh, but that would be a lot more work.

    answered Apr 18 at 10:11

    toolic's user avatar

    toolictoolic

    57.9k17 gold badges75 silver badges117 bronze badges

    0

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