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
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asked Apr 22, 2014 at 17:26
You need a semicolon at the end of the first line:
module shifter16 (A, H_sel, H);
answered Apr 22, 2014 at 17:39
AriAri
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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
Version Found: 13.1
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.
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.
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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
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
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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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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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
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
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Thread starterGaurav Sarode
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Start dateMay 9, 2012
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Code Verilog - [expand] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 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
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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
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#4
You are trying to instantiate modules inside an always block. That’s not possible according to Verilog rules. (Neither in VHDL, by the way).
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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
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#6
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.
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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
here in first table H stands for population of frequency of that perticular RGB combination.
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In MATLAB it was done with two nested for loops…
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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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— — — Updated — — —
hey thnx FvM n atulaxc i got the sequential accessing thing…
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