Bison Grammar %type and %token - grammar

Question

Why is it that I have to use $<nVal>4 explicitly in the below grammar snippet?
I thought the %type <nVal> expr line would remove the need so that I can simply put $4?
Is it not possible to use a different definition for expr so that I can?
%union
{
int nVal;
char *pszVal;
}
%token <nVal> tkNUMBER
%token <pszVal> tkIDENT
%type <nVal> expr
%%
for_statement : tkFOR
tkIDENT { printf( "I:%s\n", $2 ); }
tkEQUALS
expr { printf( "A:%d\n", $<nVal>4 ); } // Why not just $4?
tkTO
expr { printf( "B:%d\n", $<nVal>6 ); } // Why not just $6?
step-statement
list
next-statement;
expr : tkNUMBER { $$ = $1; }
;
Update following rici's answer. This now works a treat:
for_statement : tkFOR
tkIDENT { printf( "I:%s\n", $2 ); }
tkEQUALS
expr { printf( "A:%d\n", $5 /* $<nVal>5 */ ); }
tkTO
expr { printf( "A:%d\n", $8 /* $<nVal>8 */ ); }
step-statement
list
next-statement;

Answer 1

Why is it that I have to use $<nVal>4 explicitly in the below grammar snippet?
Actually, you should use $5 if you want to refer to the expr. $4 is the tkEQUALS, which has no declared type, so any use must be explicitly typed. $3 is the previous midrule action, which has no value since $$ is not assigned in that action.
By the same logic, the second expr is $8; $6 is the second midrule action, which also has no value (and no type).
See the Bison manual:
The mid-rule action itself counts as one of the components of the rule. This makes a difference when there is another action later in the same rule (and usually there is another at the end): you have to count the actions along with the symbols when working out which number n to use in $n.