[dcl.type.auto.deduct]/2 says:
"""
A type T containing a placeholder type, and a corresponding initializer E, are determined as follows:
— for a variable declared with a type that contains a placeholder type, T is the declared type of the variable and E is the initializer. ...
"""
[dcl.type.auto.deduct]/4 says:
"""
If the placeholder-type-specifier is of the form type-constraintopt auto, the deduced type T replacing T is determined using the rules for template argument deduction.
Obtain P from T by replacing the occurrences of type-constraintopt auto either with a new invented type template parameter U, or ... [doesn't matter].
Deduce a value (!!!) for U using the rules of template argument deduction from a function call, where P is a function template parameter type and the corresponding argument is E.
...
[Example:
The type of i is the deduced type of the parameter u in the call f(expr) of the following invented function template:
template <class U> void f(const U& u);
— end example]
"""
There are several issues here (both in normative wording and the Example): the major one is that for a variable declared as const auto &i = expr;, the initializer is = expr, not expr. And one can't call a function with this initializer as an argument, like f(= expr). Probably, the paragraph should say that the initializer E is the default argument of an invented function template (and the type is deduces as if the function is called without argument)? Nah, the grammar for parameter-declaration doesn't allow initializer.
Also, is it ok that the Standard calls the deduced type "a value of a type parameter" and uses the instructing tone like Obtain P from T ... Deduce a value ....
[dcl.type.auto.deduct]/2 says:
"""
A type
Tcontaining a placeholder type, and a corresponding initializer E, are determined as follows:— for a variable declared with a type that contains a placeholder type,
Tis the declared type of the variable and E is the initializer. ..."""
[dcl.type.auto.deduct]/4 says:
"""
If the placeholder-type-specifier is of the form type-constraintopt
auto, the deduced typeTreplacingTis determined using the rules for template argument deduction.Obtain
PfromTby replacing the occurrences of type-constraintoptautoeither with a new invented type template parameterU, or ... [doesn't matter].Deduce a value (!!!) for
Uusing the rules of template argument deduction from a function call, wherePis a function template parameter type and the corresponding argument is E....
[Example:
The type of
iis the deduced type of the parameteruin the callf(expr)of the following invented function template:— end example]
"""
There are several issues here (both in normative wording and the Example): the major one is that for a variable declared as
const auto &i = expr;, the initializer is= expr, notexpr. And one can't call a function with this initializer as an argument, likef(= expr).Probably, the paragraph should say that the initializer E is the default argument of an invented function template (and the type is deduces as if the function is called without argument)?Nah, the grammar for parameter-declaration doesn't allow initializer.Also, is it ok that the Standard calls the deduced type "a value of a type parameter" and uses the instructing tone like Obtain
PfromT... Deduce a value ....