[dcl.type.auto.deduct] Clarify initializer for placeholder type deduction.

Author: jensmaurer

source/declarations.tex

@@ -1923,25 +1923,28 @@
 are determined as follows:
 \begin{itemize}
 \item
-for a non-discarded \tcode{return} statement that occurs
+For a non-discarded \tcode{return} statement that occurs
 in a function declared with a return type
 that contains a placeholder type,
 \tcode{T} is the declared return type
 and $E$ is the operand of the \tcode{return} statement.
 If the \tcode{return} statement
 has no operand,
-then $E$ is \tcode{void()};
+then $E$ is \tcode{void()}.
 \item
-for a variable declared with a type
+For a variable declared with a type
 that contains a placeholder type,
-\tcode{T} is the declared type of the variable
-and $E$ is the initializer.
+\tcode{T} is the declared type of the variable.
+If the initializer of the variable is a \grammarterm{brace-or-equal-initializer}
+of the form \tcode{= \grammarterm{initializer-clause}},
+$E$ is the \grammarterm{initializer-clause}.
 If the initialization is direct-list-initialization,
 the initializer shall be a \grammarterm{braced-init-list}
 containing only a single \grammarterm{assignment-expression}
-and $E$ is the \grammarterm{assignment-expression};
+and $E$ is the \grammarterm{assignment-expression}.
+Otherwise, $E$ is the initializer.
 \item
-for a non-type template parameter declared with a type
+For a non-type template parameter declared with a type
 that contains a placeholder type,
 \tcode{T} is the declared type of the non-type template parameter
 and $E$ is the corresponding template argument.