How to convert 8 bits to 16 bits in VHDL?

architecture RTL of test is
    signal s8: std_logic_vector(7 downto 0);
    signal s16: std_logic_vector(15 downto 0);
begin
    s16 <= X"00" & s8;
end;

For completeness, yet another way which is occasionally useful:

--  Clear all the slv_16 bits first and then copy in the bits you need.  
process (slv_8)
begin
    slv_16 <= (others => '0');
    slv_16(7 downto 0) <= slv_8;
end process;

I've not had to do this for vectors that I can recall, but I have had need of this under more complex circumstances: copying just a few relevant signals into a bigger, more complex, record was one time.

This handles the conversion without having to edit the widths of the zeroes if either std_logic_vector changes:

architecture RTL of test is
    signal s8: std_logic_vector(7 downto 0);
    signal s16: std_logic_vector(15 downto 0) := (others => '0');
begin
    s16(s8'range) <= s8;
end;

If the 8 bit value is interpreted as signed (2's complement), then the general and standard VHDL conversion method is to use the IEEE numeric_std library:

library ieee;
use ieee.numeric_std.all;

architecture sim of tb is
    signal slv_8  : std_logic_vector( 8 - 1 downto 0);
    signal slv_16 : std_logic_vector(16 - 1 downto 0);
begin
    slv_16 <= std_logic_vector(resize(signed(slv_8), slv_16'length));
end architecture;

So first the std_logic_vector is converted to a signed value, then the resize is applied, which will sign extend the signed value, and the result is finally converted back to std_logic_vector.

The conversion is rather lengthy, but has the advantage that it is general and works even if the target length is changed later on.

The attribute 'length simply returns the length of the slv_16 std_logic_vector, thus 16.

For unsigned representation instead of signed, it can be done using unsigned instead of signed, thus with this code:

    slv_16 <= std_logic_vector(resize(unsigned(slv_8), slv_16'length));