Why is it wrong when I separately assign the output values?

Question

The following code is correct and successfully complied.

module top_module (
    input [4:0] a, b, c, d, e, f,
    output [7:0] w, x, y, z );//

    assign {w,x,y,z} = {a,b,c,d,e,f,2'b11};
    

endmodule

The next is wrong, which is confusing.

module top_module (
    input [4:0] a, b, c, d, e, f,
    output [7:0] w, x, y, z );//
   
    assign w = {a,b[2:0]};
    assign x = {c[4:3],b[2:0]};
    assign y = {c[4:3],d[2:0]};
    assign z = {d[4:3],2'b11};

endmodule

Why is it wrong when I separately assign the output values?

By the way, here's the original problem:

Given several input vectors, concatenate them together then split them up into several output vectors. There are six 5-bit input vectors: a, b, c, d, e, and f, for a total of 30 bits of input. There are four 8-bit output vectors: w, x, y, and z, for 32 bits of output. The output should be a concatenation of the input vectors followed by two 1 bits:

I've tried so many times, and I just can't assign values to them separately.

Answer 1

You need to be a little more careful when counting the number of bits in each assignment. You need to make sure the RHS of each has 8 bits. And, you need to use the proper part-selects of each input. Your code does not even use 2 of the inputs: e and f.

This is equivalent to your 1st code example:

module top_module (
  input [4:0] a, b, c, d, e, f,
    output [7:0] w, x, y, z );//
   
  assign w = {a[4:0], b[4:2]       };
  assign x = {b[1:0], c[4:0], d[4] };
  assign y = {d[3:0], e[4:1]       };
  assign z = {e[0]  , f[4:0], 2'b11};
endmodule

It is not necessary to use [4:0], but I chose to do so to make it more obvious how many bits are on the RHS.