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Error 10170 Verilog HDL Syntax Error? Repair Immediately

 

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    Over the past few days, some of our readers have informed us that they have encountered syntax error 10170 verilog hdl. / 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).

     

     

    The reason for the syntax error is that you can’t just write:

      product [7: 4] = 4'b0000; 
      Assign product [7: 4] corresponds to 4'b0000; 

    error 10170 verilog hdl syntax error

    But if you’re not using System Verilog (and your old-fashioned coding style assumes clients don’t), you’ll find that most

      assign product [7: 4] = 4'b0000; 

    Also,
    error 10170 verilog hdl syntax error

    doesn’t compile because the target of assign should be wire , not the most recent reg . And if you change the product to a full transfer , you will find that these transactions give you an error:

      product = model 1; >> // move to the right and assign 0 to my high orderProduct [7: 3] means product [7: 3] + multiplicand [4: 0]; // add 5 bits so we can control the report 
      product = part 1; >> // move right 

    because a person cannot match thread in a large always (or initial ) block.

    You always seem to design some kind of offset and add a multiplier and / or you probably want to initialize the product at the beginning of the calculation. (Assuming you are grouping syntax) lines

      (assign) product or service [7: 4] 4'b0000;(Assign) = Product [3: 0] = Multiplier [3: 0]; 

    Updated

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    continuous and continuous promotion of the product ; they do not initialize product . This is where you design personal computers, not write software.

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     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 every time clock)beginif (reset)q (q1 == ~ q1)beginalways @ (Posege <= 4'b0000;otherwise, if (q <4'b0011)d <= d + 1;anothera = d [2] and d [3];endendFinal module 

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    Error 10170 Syntax error while compiling

    I really feel like a newbie to Verilog. There is a corresponding compilation error for the if statement.
    Can anyone help me by pointing out my biggest mistake?

    This is a piece of code I have written. Yes (10170): the following
    errors

    Verilog HDL syntax error with seqdet.v (24) next to "if" text;
    Waiting for an identifier ("if" is definitely a reserved keyword) or any type of number, system task, or "(", plus "{" or unary operator,

    current_state is a kind of register, and reset_state is initialized to 3'b000, which allows the parameter instruction to be used.

    Post by Jughead
    I'm new to Verilog. System error of the if statement.
    Can anyone help me by pointing out seriousIs there a mistake? [2: 0]
    reg next_state, current_state;
    parameters reset_state = 3'b000;
    case (data)
    {
    if (current_state == reset_state)
    begin

    next_state = reset_state;

    end
    This is the last part of the code I wrote down. The following
    errors (10170): Verilog HDL syntax error in seqdet.v (24) next to the words "if";
    expects an identifier ("if" is a reserved search expression), or, or a number, a system task, or "(", or "{", or a unary operator,
    current_state may have a type case and reset_state inserts have been initialized up to 3'b000 using the parameter operator.
    Thanks,
    Aravind

    Above, I found out that the case did not have instructions on how to start the process and how to end it
    and usually they had parentheses that, in my opinion, were not needed. I'm not sure if I'm right, but I'm not getting an error right now.

    Are located used for bitstring. Use start-end instead.
    Don't forget about the "endcase".

    Are located used for bitstring. Use start-end instead.
    Don't forget, however, "endcase".

     

     

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