Error 10170 verilog hdl syntax error at near text

Working with 2014 version of Quartus II software (web edition), I receive the error 10170 when compiling the following code: module shifter16 (A, H_sel, H) input [15:0]A; input H_sel; output [15:0...

Working with 2014 version of Quartus II software (web edition), I receive the error 10170 when compiling the following code:

module shifter16 (A, H_sel, H)
input [15:0]A;
input H_sel;
output [15:0]H;
reg [15:0] H;
always @ (A or H_sel)
begin
    if (H_sel)
        H={A[14:0],1'b0};
    else
        H={A[15],A[15:1]};
end
endmodule

Error received:

Error (10170): Verilog HDL syntax error at shifter16.v(2) near text «input»; expecting «;»

Qiu's user avatar

Qiu

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asked Apr 22, 2014 at 17:26

user3561441's user avatar

You need a semicolon at the end of the first line:

module shifter16 (A, H_sel, H);

answered Apr 22, 2014 at 17:39

Ari's user avatar

AriAri

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0

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Article ID: 000075173

Content Type: Error Messages

Last Reviewed: 11/24/2014

Error (10170): Verilog HDL syntax error at <Verilog_file>.v(line_number) near text «,»; expecting an operand

Environment

Quartus Edition

  • Quartus® II Subscription Edition
  • Version Found: 13.1

    BUILT IN — ARTICLE INTRO SECOND COMPONENT

    Description

    Due to a problem in the Quartus® II software version 13.1 and later, you may get the following error when compiling a Verilog HDL  file that has converted from a Block Design File (.bdf).

    The cause of the error is due to the generated Verilog HDL file has a extra comma in the port connections.

    Resolution

    To workaround the error, manually delete the extra comma in the <Verilog_file>.v(line_number).

    This problem is schedule to be fixed in future release of the Quartus II software.

    • Description
    • Resolution

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    Article ID: 000074408

    Content Type: Error Messages

    Last Reviewed: 06/18/2007

    Error (10170): Verilog HDL syntax error at <location> near text «generate»; expecting «end», or an identifier («generate» is a reserved keyword ), or a sequential statement

    Environment

    BUILT IN — ARTICLE INTRO SECOND COMPONENT

    Description

    You may get this error if your design uses extra generate/endgenerate statements for nested loops. Older versions of the Quartus® II software erroneously accepted nested generate/endgenerate statements in Verilog HDL design files. The Quartus II software beginning with version 6.0 correctly flags nested generate/endgenerate statements as an error.

    If you have loops within a loop, you only need one generate/endgenerate pair, as shown in the following example that reverses the bits in a bus.

    genvar i,j;
    generate
    for( i=0; i<8; i=i 1 )
    begin : outer
        for (j=0;  j<8; j=j 1 )
            begin : inner
            assign data_out[i][j] = data_in[7-i][7-j]; 
        end
    end
    endgenerate

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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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    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».

    Speed up your PC today with this easy-to-use download.

    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?


    • Вопрос задан

      более трёх лет назад

    • 1392 просмотра

    Пригласить эксперта

    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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    Article ID: 000079408

    Content Type: Error Messages

    Last Reviewed: 09/11/2012

    Error (10170): Verilog HDL Syntax Error at <filename> near text «int»; expecting an identifier («int» is a reserved keyword)

    Environment

    BUILT IN — ARTICLE INTRO SECOND COMPONENT

    Description

    In the Quartus® II software may generate this error when you declare multiple loop variables within a SystemVerilog FOR loop, because this syntax is currently unsupported.

    The following is an example of unsupported syntax:
    for(int i=0, int j=0; i<4, j<2; i , j )

    To work around this problem, declare only one loop variable within the FOR loop.

    Use the following supported syntax to work around this problem:
    int j=0;
    for(int i=0, j=0; i<4, j<2; i , j )

    This problem is scheduled to be fixed in a future release of the Quartus II software.

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    Error in Verilog problem


    • Thread starter

      Gaurav Sarode


    • Start date

      May 9, 2012

    Status
    Not open for further replies.

    • #1

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    Hello everyone I am implementing Image segmentation on FPGA, for that i have to compare the pixel values with each other. I have stored image pixel values in ROM, and now want to put them in RAM and simultaneously compare them with previous entries in RAM.I have written this code given bellow but getting lot of errors. So please help


    Code Verilog - [expand]
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    module imp(clk, addr, we_f, en_f, do_r, do_g, do_b, do_t1, do_t2);
     
    input clk; input en; input we;
     
    output [10:0]do_r[483:0]; 
    output [10:0]do_g[483:0];
    output [10:0]do_b[483:0];
    output [10:0]do_t1[483:0];
    output [10:0]do_t2[483:0];
     
    reg [10:0]temp_r;
    reg [10:0]temp_g;
    reg [10:0]temp_b;
    reg [10:0]addr;
    reg [10:0]r1=10'b0000000001;
    reg [10:0]r2;
    reg decision;
    reg cn1;
    reg cn2;
     
    // prcedural loop for getting bins from ROM & processiing them
     
    always @(posedge clk)
     
    begin
        for (addr=0; addr<=484; addr=addr+1)
        
        rdrom gt1(clk, en_f, addr, temp_r);
        grom gt2(clk, en_f, addr, temp_g);
        brom gt3(clk, en_f, addr, temp_b);
        
            if (addr==0)
            begin
            ram_r gt4(clk, we_f, en_f, addr, temp_r, do_r);
            ram_g gt5(clk, we_f, en_f, addr, temp_g, do_g);
            ram_b gt6(clk, we_f, en_f, addr, temp_b, do_b );
            ram_t1 gt7(clk, we_f, en_f, addr, r1, do_t1 );
            ram_t2 gt8(clk, we_f, en_f, addr, r1, do_t2 );
            break;
            end
        
            if(ram_r[addr]==0 && ram_g[addr]==0 && ram_b[addr]==0 )
        
            begin
            r2<= addr;
        
            for(i=0; i<r2; i=i+1)
            begin
            segmtnbox gt9(temp_r, temp_g, temp_b, RAM_r[i], RAM_g[i], RAM_b[i], decision, w9s, w11dif, cin);
        
                if(decision==1)
                begin
                    ram_t2 gt10(clk, we_f, en_f, addr, i, d_t2);
                    RAM_t1[addr]=RAM_t1[addr]+1;
            
                    break;
                    
                end
        
                cn1=1;
        
            end
            
            if(cn1==1)
            begin
            
                ram_r gt11(clk, we_f, en_f, addr, temp_r, do_r);
                ram_g gt12(clk, we_f, en_f, addr, temp_g, do_g);
                ram_b gt13(clk, we_f, en_f, addr, temp_b, do_b);
                ram_t1 gt14(clk, we_f, en_f, addr, r1, do_t1 );
                ram_t2 gt15(clk, we_f, en_f, addr, r1, do_t2 );
                
            end
        
        end

    Last edited by a moderator: May 9, 2012

    • #2

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    There is an error in input/output declaration. You have used «we» instead of «we_f» and «en» instead of «en_f». You have used internal bus «addr» which is of same name as the port name. We cannot use internal signals and ports with the same name.

    • #3

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    @ Akhilkumar
    Thanx for ur rpl

    I have made changes that u suggested but still m getting errors, m using Quartus II

    the errors are as follows
    Error (10170): Verilog HDL syntax error at imp.v(28) near text «gt1»; expecting «<=», or «=»
    Error (10170): Verilog HDL syntax error at imp.v(29) near text «gt2»; expecting «<=», or «=»
    Error (10170): Verilog HDL syntax error at imp.v(30) near text «gt3»; expecting «<=», or «=»
    Error (10170): Verilog HDL syntax error at imp.v(34) near text «(«; expecting «;», or «,»
    Error (10149): Verilog HDL Declaration error at imp.v(34): identifier «temp_r» is already declared in the present scope
    Error (10170): Verilog HDL syntax error at imp.v(34) near text «)»; expecting «;», or «,»
    Error (10170): Verilog HDL syntax error at imp.v(35) near text «(«; expecting «;», or «,»
    Error (10149): Verilog HDL Declaration error at imp.v(35): identifier «temp_g» is already declared in the present scope
    Error (10170): Verilog HDL syntax error at imp.v(35) near text «)»; expecting «;», or «,»
    Error (10170): Verilog HDL syntax error at imp.v(36) near text «(«; expecting «;», or «,»
    Error (10149): Verilog HDL Declaration error at imp.v(36): identifier «temp_b» is already declared in the present scope
    Error (10170): Verilog HDL syntax error at imp.v(36) near text «)»; expecting «;», or «,»
    Error (10170): Verilog HDL syntax error at imp.v(37) near text «(«; expecting «;», or «,»
    Error (10149): Verilog HDL Declaration error at imp.v(37): identifier «r1» is already declared in the present scope
    Error (10170): Verilog HDL syntax error at imp.v(37) near text «)»; expecting «;», or «,»
    Error (10170): Verilog HDL syntax error at imp.v(38) near text «(«; expecting «;», or «,»
    Error (10149): Verilog HDL Declaration error at imp.v(38): identifier «r1» is already declared in the present scope
    Error (10170): Verilog HDL syntax error at imp.v(38) near text «)»; expecting «;», or «,»
    Error (10170): Verilog HDL syntax error at imp.v(49) near text «(«; expecting «;», or «,»
    Error: Quartus II Analysis & Synthesis was unsuccessful. 19 errors, 7 warnings
    Error: Peak virtual memory: 183 megabytes
    Error: Processing ended: Thu May 10 10:45:19 2012
    Error: Elapsed time: 00:00:02
    Error: Total CPU time (on all processors): 00:00:02
    Error: Quartus II Full Compilation was unsuccessful. 21 errors, 7 warnings

    ———- Post added at 06:23 ———- Previous post was at 06:17 ———-

    I have written code fro ROM where i hve stored image information and have separate code for RAM which is being modified depending on the operation carried out by sgmtnbox i.e. depends on its output decison

    • #4

    A general suggestion first. Get you a Verilog text book to understand about basic requirements of module design. Or review the Verilog template in Quartus code editor, accessible by a right mouse click.

    You are trying to instantiate modules inside an always block. That’s not possible according to Verilog rules. (Neither in VHDL, by the way).

    • #5

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    @Fvm

    I want to implement iterative operation on FPGA…and m not getting a way out of it pls help

    ———- Post added at 08:00 ———- Previous post was at 07:59 ———-

    @Fvm

    I want to implement iterative operation on FPGA…and m not getting a way out of it pls help

    • #6

    I fear, you have to go back to basic HDL design concepts first.

    A for loop in HDL is not defining a sequence in time rather than parallel logic. Apart from other problems, the iteration construct in your code will generate 485 logic instances, which very likely execeeds any reasonable resource limits.

    Module instances have to placed in concurrent code, outside laways blocks. To generate multiple instances in a loop, use a generate construct.

    I guess, that th eintended iteration can be implemented as sequential processing, one iteration step per clock cycle. But I don’t understand th eapplication problem in detail.

    • #7

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    @FvMThanx for reply

    I want to perform image segmentation for that i have to compare the pixel color value with all the other pixels color values, so i have saved the pixel color values in ROM and want to get following two tables from it

    tables.png here in first table H stands for population of frequency of that perticular RGB combination.

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    I have implemented it on MATLAB n now trying it on FPGA. Because repetitive comparison operation in algorithm m facing problem in it.
    In MATLAB it was done with two nested for loops…

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    Is there any way to use designed hardware for operating on matrix saved in RAM on FPGA…

    • #10

    Data in RAM has to be accessed sequentially, one address per clock cycle.

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    One of the good Verilog Tutorials I know will answer most of your basic Verilog questions. Its explained with many examples.

    https://www.fullchipdesign.com/verilog_tutorial.htm

    Synchronous Random Access Memory (RAM) implementation in verilog is at following link.

    **broken link removed**

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    I want to store N*3 matrix on FPGA kit RAM i.e. dont want to geenerate hardware for it. and den access these matrix rows one by one pls help

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    With reference to sync_memory explanation on **broken link removed** You can infer RAM on fpgas by just removing the reset condition from the always block.**broken link removed** the memory is going to be inferred using logic cells.

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    hey thnx FvM n atulaxc i got the sequential accessing thing…

    — — — Updated — — —

    hey thnx FvM n atulaxc i got the sequential accessing thing…

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