CODEWITHSUNDEEP
vhdlghdl
pitazzo
pitazzo

Reputation: 1212

Bad character in identifier in GHDL

I'm trying to compile the following component in GHDL in Ubuntu 17.10. This is the code of the component:

library ieee;
use ieee.std_logic_1164.all;
use ieee.numeric_std.all; -- Unsigned

entity simple is end simple;
architecture behaviour of simple is
    signal clk : std_logic := '0';
    signal sigterm : std_logic := '0';
    signal counter : unsigned(7 downto 0) := x"00";
begin
    process
    begin
        wait for 5 us;
        clkloop : loop
            wait for 1 us;
            clk <= not clk;
            if sigterm = '1' then
                exit;
            end if;
        end loop clkloop;
        wait for 5 us;
        wait;
    end process;

    process (clk)
    begin
        if rising_edge(clk) then
            if counter = 16 then
                sigterm <= '1';
            end if;
            counter <= counter + 1;
        end if;
    end process;
end behaviour;

I'm getting this error message while elaborating the component:

ghdl:*command-line*: bad character in identifier

I'm using the following flags in order to make the commands work in my 64-bits machine, I don't know if they are totally OK:

ghdl -a -Wa,--32 -Wl,--32 simple.vhdl
ghdl -e -Wa,--32 -Wl,--32 simple.vhdl

Upvotes: 1

Views: 1583

Answers (1)

Renaud Pacalet
Renaud Pacalet

Reputation: 28945

This is an elaboration error message: what is expected here is the name of a design unit, not a file name, plus, if it is not the default work, the name of the library.

ghdl -a -Wa,--32 -Wl,--32 simple.vhdl
ghdl -e -Wa,--32 -Wl,--32 simple

Or, if you compile simple.vhdl in library foolib:

ghdl -a -Wa,--32 -Wl,--32 --work=foolib simple.vhdl
ghdl -e -Wa,--32 -Wl,--32 --work=foolib simple

Upvotes: 1

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