Vhdl syntax error near for

Vhdl syntax error near begin

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Error 10500: VHDL Syntax

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I am working on a code for a basic vending machine that will give out a product at 75cents

BUt for some reason I dont know how to fix this syntax error

Library ieee; Use ieee.numeric_std.all; Use ieee.std_logic_1164.all; Use ieee.std_logic_unsigned.all; Entity Lab06 is Port ( Clock :IN std_logic; Reset_n :IN std_logic; Quarter_in :IN std_logic; Dime_in :IN std_logic; Nickel_in :IN std_logic; Pennny_in :IN std_logic; Coin_Return :IN std_logic); End Lab06; Architecture Behavior Of Lab06 is Signal int_counter: std_logic_vector(25 downto 0); Constant MAXVALUE: std_logic_vector(25 downto 0):= «10111110101111000010000000»; —One Second Count ————————————————————————————————- Type State_type is (wait1, penny, dime, nickel, quarter, enough, excess, vend, change); Signal current_state,next_state: state_type; Signal Money: std_logic_vector(6 downto 0); Signal Red_Bull: std_logic; Signal Change_Back: std_logic; Begin Counter_Clock:Process (clock, reset_n) Begin If reset_n = ‘0’ Then Int_counter ‘0’); Elsif (rising_edge(clock)) Then If int_counter = MAXVALUE Then int_counter ‘0’); Else int_counter If (money = «0000000»)Then — No money in vending machine next_state = «1001011»)Then next_state Next_state Next_state Next_state Next_state If (money >= «1001011»)Then Next_state Next_state Next_state Next_state next_state Money Money Money Money Money Money Money Money Money Money

The errors I am getting are

Info: Running Quartus II 32-bit Analysis & Synthesis

Info: Version 12.0 Build 178 05/31/2012 SJ Full Version

Info: Processing started: Wed Oct 31 15:32:29 2012

Info: Command: quartus_map —read_settings_files=on —write_settings_files=off Lab06 -c Lab06

Info (20030): Parallel compilation is enabled and will use 4 of the 4 processors detected

Info (12021): Found 2 design units, including 1 entities, in source file hex_display.vhd

Info (12022): Found design unit 1: Hex_Display-Structure

Info (12023): Found entity 1: Hex_Display

Error (10500): VHDL syntax error at Lab06.vhd(108) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(111) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(114) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(117) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(120) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(126) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(128) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(130) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(132) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement

Error (10500): VHDL syntax error at Lab06.vhd(134) near text «Case»; expecting «if»

Error (10500): VHDL syntax error at Lab06.vhd(136) near text «Process»; expecting «if»

Error (10500): VHDL syntax error at Lab06.vhd(139) near text «Begin»; expecting «:=», or » 0 баллов

Источник

Why I have If-then VHDL errors in my code?

Opel_Corsa

Member level 1

epoxi

Newbie level 2

mmh, strange, I’ll check that

bakkali

Junior Member level 1

if else in vhdl

first i think that you miss a «end if» in your code
you have two if so you mast write two end if
i wish to view all the code if you can

Girisha

Newbie level 2

vhdl if statement outside process

U have write sequential statements inside the process.
The if else statements are sequential statements. so u need to write these statements inside the process.
I think u r trying to write state mechine it should be change the state with respect to clock,
include clock event statements in process sensitivity list, u can refer any VHDL state mechine examples.

rsrinivas

Advanced Member level 1

vhdl if then else

if(m==load) is the correct syntax
if(condition)
u r using
if(assignment)
as girisha told make sure ur if part is in a process

lucbra

Advanced Member level 2

vhdl illegal sequential statement

There is missing a begin statement after the type and signal declarations.

Opel_Corsa

Member level 1

error (10500) vhdl

Thanks very much for all the responses. First, the code that I posted was not a state machine. My code does deal with a state machine, but the one posted wasn’t one. Also there are correct number of «end if» statements in my code. In any case, here’s the full part of the architecture (the code is still incomplete, but I think it should compile nevertheless):

Elephantus

Junior Member level 3

Is a sequential statement, not a concurrent one, and cannot be written outside a process begin/end block. Therefore, you can write it as:

Or, you might use the concurrent assignment form:

which can be safely written in an architecture body.

Hope this helps.

Opel_Corsa

Member level 1

vhdl if condition

Thanks. Can you explain why you are considering m to be a clock? (since the if-then clause is sequential). I thought it was purely combinational.

rsrinivas

Advanced Member level 1

syntax of if statement in vhdl

if is a sequential statement and it should be present in a process.
But when u synthesise if statements it becomes MUX which is combinational.

Elephantus

Junior Member level 3

error (10500): vhdl

In this process, m is not a clock. Synchronous (clocked) logic is described in a different manner in VHDL.

When you define a process:

The signals in the parentheses are the members of the process’ sensitivity list. This means that the sequential statements in the process will re-execute and re-evaluate on any change of any signal in the list. The code described above is a purely combinatorial process:
The general rule is: when you describe complex combinatorial functions in VHDL, use processes and assign ALL inputs of the combinatorial process to the sensitivity list.

Synchronous processes (registers) need to trigger only on a rising or falling edge of a clock signal.

The process above describes a d type flip flop which samples the input d on the rising edge of clk. Note that only the global set-reset (gsr) and the clock signal are in the process sensitivity list. That’s because these are the only signals the process must react to. As you can see, the presence of the clk signal in the sensitivity list is not sufficient to make it a clock signal. The condition that the change happens at rising_edge(clk) makes clk a clock signal, and the synthesis tools recognize clk as a clock signal through the rising edge condition.

The general rule would be: When describing synchronous logic, use processes triggered by a clk and an async reset signal (if needed), and make the assignment conditional by the rising_edge(clk) or falling_edge(clk). (see code above)

Also, don’t mix sequential statements with sequential logic. VHDL as a language consists of sequential statements (which are executed in order) and concurrent statements (which are executed in parallel). Sequential statements do not directly correspond with sequential(synchronous) logic.

Источник

I am working on a code for a basic vending machine that will give out a product at 75cents 

BUt for some reason I dont know how to fix this syntax error 


Library ieee;
Use ieee.numeric_std.all;
Use ieee.std_logic_1164.all;
Use ieee.std_logic_unsigned.all;
Entity Lab06 is
Port (
Clock :IN std_logic;
Reset_n :IN std_logic;
Quarter_in :IN std_logic;
Dime_in :IN std_logic;
Nickel_in :IN std_logic;
Pennny_in :IN std_logic;
Coin_Return :IN std_logic);
End Lab06;
Architecture Behavior Of Lab06 is
Signal int_counter: std_logic_vector(25 downto 0);
Constant MAXVALUE: std_logic_vector(25 downto 0):= "10111110101111000010000000"; --One Second Count
-------------------------------------------------------------------------------------------------
Type State_type is (wait1, penny, dime, nickel, quarter, enough, excess, vend, change);
Signal current_state,next_state: state_type;
Signal Money: std_logic_vector(6 downto 0);
Signal Red_Bull: std_logic;
Signal Change_Back: std_logic;
Begin
Counter_Clock:Process (clock, reset_n)
Begin
If reset_n = '0' Then
Int_counter <= (Others => '0');
Elsif (rising_edge(clock)) Then
If int_counter = MAXVALUE Then
int_counter <= (Others => '0');
Else
int_counter <= int_counter + '1';
End If;
End If;
End Process;
Sync:Process (clock, reset_n)
Begin
If (reset_n = '0')then
Current_state <= Wait1;--Wait1 is the default state_type
Elsif(rising_edge(clock))Then
If (int_counter = MAXVALUE)Then
Current_state <=next_state;-- Advance the state(Red Bull)machine
End If;
End If;
End Process;
Comb:Process(int_counter,current_state, Quarter_in, Dime_in, Nickel_in, Pennny_in, money, Coin_Return)
Begin
If (int_counter = MAXVALUE) Then
Case(current_state)is
When wait1=>
If (money = "0000000")Then -- No money in vending machine
next_state <= Wait1;
If (Quarter_in = '1')Then -- Money is inserted
next_state <= Quarter;
If (Dime_in = '1')Then
next_state <= Dime;
If (Nickel_in = '1')Then
next_state <= Nickel;
If (Pennny_in = '1')Then
next_state <= Penny;
If (money >= "1001011")Then
next_state <= Enough;
If(Coin_Return = '1')Then
next_state <= Change;
Else
next_state <=Wait1;
End If;
----------------------------------------------------------------- After Quarter Inserted, next step
When Quarter =>
Next_state <= Wait1;
----------------------------------------------------------------- After Dime Inserted, next step
When Dime =>
Next_state <= Wait1;
----------------------------------------------------------------- After Nickel Inserted, next step
When Nickel =>
Next_state <= Wait1;
----------------------------------------------------------------- After Penny Inserted, next step
When Penny =>
Next_state <= Wait1;

When Enough =>
If (money >= "1001011")Then
Next_state <= Excess;
Else
Next_state <= vend;
End If;
When Excess =>
Next_state <= vend;
When vend =>
Next_state <= Wait1;
When Change =>
Next_state <= Wait1;
When OTHERS =>
next_state <= Wait1;
End Case;
End If;
End Process;
----------------------------------------------------------------- Money calculation
Money_Calc:Process(Current_state, money)
Begin
Case (current_state) is
When wait1 =>
Money <= Money;
When Quarter =>
Money <= Money + "0011001";
When Dime =>
Money <= Money + "0001010";
When Nickel =>
Money <= Money + "0000101";
When Penny =>
Money <= Money + "0000001";
When Enough =>
Money <= money;
When Excess =>
Money <= Money;
When vend =>
Money <= Money - "1001011";
When change =>
Money <= "0000000";
When OTHERS =>
Money <= Money;
End Case;
End Process;
End Behavior;


The errors I am getting are 

Info: ******************************************************************* 

Info: Running Quartus II 32-bit Analysis & Synthesis 

Info: Version 12.0 Build 178 05/31/2012 SJ Full Version 

Info: Processing started: Wed Oct 31 15:32:29 2012 

Info: Command: quartus_map —read_settings_files=on —write_settings_files=off Lab06 -c Lab06 

Info (20030): Parallel compilation is enabled and will use 4 of the 4 processors detected 

Info (12021): Found 2 design units, including 1 entities, in source file hex_display.vhd 

Info (12022): Found design unit 1: Hex_Display-Structure 

Info (12023): Found entity 1: Hex_Display 

Error (10500): VHDL syntax error at Lab06.vhd(108) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(111) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(114) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(117) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(120) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(126) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(128) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(130) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(132) near text «When»; expecting «end», or «(«, or an identifier («when» is a reserved keyword), or a sequential statement 

Error (10500): VHDL syntax error at Lab06.vhd(134) near text «Case»; expecting «if» 

Error (10500): VHDL syntax error at Lab06.vhd(136) near text «Process»; expecting «if» 

Error (10500): VHDL syntax error at Lab06.vhd(139) near text «Begin»; expecting «:=», or «<=» 

Error (10500): VHDL syntax error at Lab06.vhd(162) near text «Process»; expecting «if» 

Info (12021): Found 0 design units, including 0 entities, in source file lab06.vhd 

Error: Quartus II 32-bit Analysis & Synthesis was unsuccessful. 13 errors, 0 warnings 

Any assistance would be very helpful.

I’m a trying to find out the problem with this simple VHDL code. I’ll be grateful if someone could help me. PS: I tried the code without the conditional block and it works :S !

*The message error is : Error (10500): VHDL syntax error at Four_Bits_Adder.vhd(18) near text «if»; expecting «;», or an identifier («if» is a reserved keyword), or «architecture»*

The 4 bits adder code is :

Library ieee;
Use ieee.std_logic_1164.all;
Use ieee.numeric_std.all;
Use ieee.std_logic_unsigned.all;

entity Four_Bits_Adder is
port(A,B:in std_logic_vector(3 downto 0);
    S:out std_logic_vector(3 downto 0);
    Cout:out std_logic);
end Four_Bits_Adder;
architecture description of Four_Bits_Adder is
begin
S<= A+B;
if A(3)=1 then
    if B(3)=1 then
        Cout<=1;
        end if;
end if;
end description;

An if statement is a sequential statement, and from your usage should either be in a process statement or you should instead use a concurrent signal assignment statement. It hardly seems complex enough for a process statement.

Concurrent assignment statement:

 Cout <= A(3) and B(3);

Note that concurrent assignment statement using an expression as the value in it’s signal assignment can pass values other than ‘0’ or ‘1’.

A different style of concurrent signal assignment statement could pass only binary representing enumeration values:

Cout <= '1'  when A(3) = '1' and B(3) = '1' else '0';

(Your if statement also appears to infer a latch and could be optimized as a constant ‘1’).

Note also that your original if nested statements could be expressed with a BOOLEAN and which is a short-circuit operator taking the place of Ada’s if expression and then .... Without one or more intervening statements following an if statement (including it’s end if) there is no reason to nest if statements with different expressions. The short-circuit operator would only evaluate the subsequent expression if the first expression evaluated true.

The form would be along the lines of

if A(3) = '1' and B(3) = '1' then
    Cout <= '1';
end if;

And could still only be used where a sequential statement is appropriate.

Note that std_logic requires enumeration values (‘U’, ‘X’, ‘0’, ‘1’,…) while 1 is a numeric value and would result in errors.