Translating a VHDL code to Verilog compilation error - vector

Question

I'm translating a VHDL code to Verilog but I have a question in VHDL:
What is the use of the concatenation with the empty string in these lines?
Xp_m5b0 <= XX_m5(23 downto 0) & "";
Yp_m5b0 <= YY_m5(23 downto 0) & "";
It is said that it changes the type, but the types here are the same (std_logic_vector).
Here are the lines that showed the type:
entity IntMultiplier_LogicOnly_24_24_48_unsigned_F400_uid4 is
port ( clk, rst : in std_logic;
X : in std_logic_vector(23 downto 0);
Y : in std_logic_vector(23 downto 0);
R : out std_logic_vector(47 downto 0) );
end entity;
signal XX_m5 : std_logic_vector(23 downto 0);
signal YY_m5 : std_logic_vector(23 downto 0);
signal Xp_m5b0 : std_logic_vector(23 downto 0);
signal Yp_m5b0 : std_logic_vector(23 downto 0);
XX_m5 <= X ;
YY_m5 <= Y ;
In verilog after translation, this concatenation gives a compilation error:
assign Xp_m5b0 = {XX_m5[23:0], 0'b };
assign Yp_m5b0 = {YY_m5[23:0], 0'b };
So does it have a difference in the meaning if I removed it and made it like this:
assign Xp_m5b0 = XX_m5[23:0];
assign Yp_m5b0 = YY_m5[23:0];

Answer 1

"" is not an empty string, but an empty array. I haven't seen it used in this context, but it can be used to convert a literal to an array. I.e. consider the next code:
entity e is end entity;
library ieee;
architecture a of e is
use ieee.std_logic_1164.all;
signal a : std_logic_vector(0 downto 0);
signal b : std_logic;
begin
-- a <= b; -- fails
a <= b&""; -- works
end architecture;
But since XX_m5(23 downto 0) is already an array (slice), it should not be required here...