About evaluation in Block

Question

I'm completely puzzled by the following codes:

Clear[func, f]
a = x^2 + z;
func[k_] := Block[{f}, (f[x_] := Evaluate[(a /. {z -> k})]; f[1])]
func[2]

2+x^2

But I think it should be 3, i.e., 2+1^2=3.

Actually for the following code:

 f[x_] := Evaluate[(a /. {z -> k})]; f[1]

The output is 3.

Can you explain the evaluation inside the Block?

Taking Simon Woods' advice, for

func[k_] := 
 Block[{f, tmp}, (tmp = Evaluate[a /. z -> k]; f[x_] = tmp; f[1])]

Trace[func[2]]

gives,

{func[2],Block[{f,tmp},tmp = Evaluate[a /. z -> 2],f[x_] = tmp; f[1]],...}

But, for the codes

func[k_] := Block[{f}, (f[x_] = Evaluate[a /. z -> k]; f[1])]
Trace[func[2]]

gives,

{func[2],Block[{f},f[x$_] = Evaluate[a /. z -> 2];f[1],...}

I notice that the arguments of f inside the Block are different in these two cases, and that is the reason.

But why is the difference?

Answer 1

Nested scoping constructs like Module or SetDelayed apply a variable renaming, adding the symbol ($) to each variable. For that, take a look at the internal form of f, as you defined it

func[k_] := Block[{f}, 
                  f[x_] := Evaluate[ReplaceAll[a, {z -> k}]];
                  Information[f];
                  Trace[f[1]]]

func[2]

Global`f
f[x$_]:=2+x^2

{f[1], 2+x^2}

the function f accepts any argument and gives the name x\$ to it. Then goes to evaluate the right-hand side and looks for any occurrence of x\$ so to make the replacement with 1, in that case. Unfortunately, the global variable a (that defined externally) depends on x and not on x$ so no replacement happens. A possible solution is to change the name of the variable x so to avoid confusion and make clear the replacement of such variable inside the ReplaceAll.

func[k_] := Block[{f},
                   f[var_] := Evaluate[ReplaceAll[a, {z -> k, x -> var}]];
                   Information[f];
                   Trace[f[1]]]


func[2]
Global`f
f[var$_]:=2+var$^2

{f[1], 2+1^2}, {1^2, 1}, 2+1, 3}