CODEWITHSUNDEEP
system-verilog
jarno
jarno

Reputation: 149

How to use let with typedef signals?

I'm trying to use let-constructs instead of `define in my TB to access some signals via hierarchical paths. It's not going that great. "Normal" signals seems to work and I can access them, but typedef-signals act strange.

Here's a super simple example code that I use to trigger the error:

module dut();

  // Create a type
  typedef struct {
    int foo;
    int bar;
  } ty_fooBar;

  // Instantiate the type
  ty_fooBar fooBar;

  // Create an "alias" for fooBar
  let fooBar2 = fooBar;

  // Assign some values and try to access the struct members
  initial begin
    fooBar.foo = 3;
    fooBar.bar = 7;
    $display("fooBar: %p",  fooBar  );
    $display("fooBar2: %p", fooBar2 );

    $display("fooBar.fooBar: %p",  fooBar.foo  );
    // $display("fooBar2.fooBar: %p", fooBar2.foo );    <- ERROR

  end

endmodule

Simulation gives this result:

# fooBar: '{foo:3, bar:7}
# fooBar2: '{foo:3, bar:7}
# fooBar.fooBar: 3

So, fooBar should now be the same as fooBar2, ModelSim shows this with the $display command, but for some reason I can access fooBar.foo but not fooBar2.foo. I tried reading the IEEE standard but that didn't enlighten me.

What's the deal with let and typedef? Do I have some profound misunderstanding?

Upvotes: 0

Views: 515

Answers (1)

dave_59
dave_59

Reputation: 42698

The let statement combines the flexibility of a `define macro with the well formed structure of function.

Like a macro. the let arguments get replaced into the body of its definition. they may be typeless.

Like a function, a let declaration is local to a scope, including a package that can be imported. References to identifiers that are not arguments are searched from the point of the declaration. And finally the problem that you are facing is that a let statement can only be called where an expression is allowed. In fact it places parenthesis around the body before substituting.

So your reference to fooBar2.foo gets expanded as (fooBar).foo which is not legal.

Upvotes: 2

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