Near End Syntax Error Unexpected End Verilog
Contents |
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies verilog syntax error near = of this site About Us Learn more about Stack Overflow the company near syntax error unexpected Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges near always syntax error unexpected always Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: verilog syntax error i give up Sign up Syntax error in Testbench file up vote 1 down vote favorite I'm trying to create a testbench file for the sequential circuit in Modelsim (verilog). But I'm getting the following syntax error. ** Error: (vlog-13069) /Assignment_2x2_tb.v(6): near "initial": syntax error, unexpected initial, expecting ';' or ','. Here's my code module seq_circuit1_tb; reg x,clk; wire q; seq_circuit1 seqct(x, clk, Q0, Q1)
Near Eof Syntax Error Unexpected End Of Source Code Verilog
//Module to generate clock with period 10 time units initial begin forever begin clk=0; #10 clk=1; #10 clk=0; end end initial begin x=0; #50 x=0; #50 x=1; #50 x=1; #50 end endmodule can anybody tell me why I'm getting this error. verilog share|improve this question edited Sep 8 '15 at 12:40 toolic 30.6k43468 asked Sep 6 '15 at 3:18 Atinesh 309418 add a comment| 2 Answers 2 active oldest votes up vote 2 down vote accepted You need a semicolon (;) after the line seq_circuit1 seqct(x, clk, Q0, Q1). share|improve this answer answered Sep 6 '15 at 3:23 MikeCAT 25.8k92746 Getting new error near "end": syntax error, unexpected end. at Line 24 just before endmodule. –Atinesh Sep 6 '15 at 3:27 #50 just before end seems to be invalid. Maybe you should add $finish; or something before the end –MikeCAT Sep 6 '15 at 3:37 add a comment| up vote 1 down vote The initial block cannot end with a delay. You need to have some statement after the last #50 as follows initial begin x=0; #50 x=0; #50 x=1; #5
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring
Syntax Error Near "always"
developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question near module syntax error verilog x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; syntax error in verilog it only takes a minute: Sign up verilog compiler syntax error unexpected end up vote 0 down vote favorite I am writing a test bench for AND module but it gives me the following error near "end": syntax error, http://stackoverflow.com/questions/32419693/syntax-error-in-testbench-file unexpected end. here is my code: module TestAND(); reg A; reg B; wire C; AND inst(A,B,C); initial begin A=1; B=0; #100 end verilog share|improve this question asked Oct 15 at 18:58 JoeEhab12 63 do #100;, instead of #100 –Karan Shah Oct 17 at 4:05 add a comment| 2 Answers 2 active oldest votes up vote 1 down vote You need to add ;after #100 and you also missing endmodule at the end. share|improve this answer answered Oct 15 http://stackoverflow.com/questions/40063018/verilog-compiler-syntax-error-unexpected-end at 19:17 Kamil Rymarz 1969 add a comment| up vote 0 down vote I see a couple things here, you have an incorrect data type: reg is only used in an always block, and so here everything should be type wire. Also if I recall correctly the primitives usually have the output first, so you would probably want AND my_and (c,a,b); However, usually you don't use primitives directly unless you're creating libraries, and would want to use assign c = a & b; instead. share|improve this answer answered Oct 16 at 11:17 maxslug 111 A reg can be assigned in an always or initial block. A wire can be assigned by an assign statement or module/primitive instantiation. Verilog primitives have lower-case names, AND could a user defined module not being shown. –Greg Oct 16 at 21:31 add a comment| Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name Email Post as a guest Name Email discard By posting your answer, you agree to the privacy policy and terms of service. Not the answer you're looking for? Browse other questions tagged verilog or ask your own question. asked 5 days ago viewed 20 times active 4 days ago Related 1Verilog compiler error0error on verilog instance?0Verilog : syntax error : unexpected SYSTEM_IDENTIFIER on us
Forum Device and Tools Related Quartus II and EDA Tools Discussion Verilog Syntax Error If this is your first visit, be sure to check out the FAQ by clicking http://www.alteraforum.com/forum/showthread.php?t=32486 the link above. You may have to register before you can post: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. Page 1 of 2 12 Last Jump to page: Results 1 to 10 of 18 Thread: Verilog Syntax Error Thread Tools Show Printable Version Email this Page… Subscribe to this syntax error Thread… Search Thread Advanced Search Display Linear Mode Switch to Hybrid Mode Switch to Threaded Mode October 30th, 2011,12:07 PM #1 Incontro View Profile View Forum Posts Altera Pupil Join Date Oct 2011 Posts 7 Rep Power 1 Verilog Syntax Error Code: module de1sign (C, SW); input [1:0] SW; output [1:0] C; assign C[0] = SW[0]; assign C[1] = SW[1]; endmodule module codes (O, C, syntax error unexpected d, e, i) ; //possibility of using "i" if "1" is reserved. input [1:0] C; output d; output e; output i; output O; endmodule begin //THIS IS LINE 17 if ( C[0] == 1'b0 && C[1] == 1'b0); O = d if ( C[0] == 1'b0 && C[1] == 1'b1); O = e if ( C[0] == 1'b1 && C[1] == 1'b0); O = i end endmodule //Line You Asked Me to Add module hexcircuit (O, HEX0); input O; output [0:6] HEX0; always @ (O); begin if ( O == d ) HEX0 = 7'b100_0010; if ( O == e ) HEX0 = 7'b011_0000; if ( O == i ) HEX0 = 7'b100_1111; end endmodule I am trying to do Lab Exercise 1 Part IV (Displaying the characters d, e, 1 on HEX0 in turn when I play with the switches.) When I try and compile it, I get: Error (10170): Verilog HDL syntax error at de1sign.v(17) near text "begin"; expecting a description I am sure this is probably a very nooby error, but help is really appreciated! Last edited by Incontro; October 30th, 2011 at 12:27 PM. Reply With Quote October 30th, 201