A cast-expression is used to explicitly convert an expression to a given
type.
cast-expression:
( type ) unary-expression
A cast-expression of the form (T)E, where T is a type and E is a
unary-expression, performs an explicit
conversion (?3.2) of the value of E to type T. If no explicit conversion
exists from the type of E to T, a
compile-time error occurs. Otherwise, the result is the value produced by
the explicit conversion. The result
is always classified as a value, even if E denotes a variable.
C# LANGUAGE SPECIFICATION
156
The grammar for a cast-expression leads to certain syntactic ambiguities.
[Example: For example, the
expression (x)?y could either be interpreted as a cast-expression (a cast
of ?y to type x) or as an additiveexpression
combined with a parenthesized-expression (which computes the value x ? y).
end example]
To resolve cast-expression ambiguities, the following rule exists: A
sequence of one or more tokens (?.4)
enclosed in parentheses is considered the start of a cast-expression only
if at least one of the following are
true:
?The sequence of tokens is correct grammar for a type, but not for an
expression.
?The sequence of tokens is correct grammar for a type, and the token
immediately following the closing
parentheses is the token ?~?, the token ?!?, the token ?(?, an identifier (?
.4.1), a literal (?.4.4), or any
keyword (?.4.3) except as and is.
The term ?correct grammar? above means only that the sequence of tokens
must conform to the particular
grammatical production. It specifically does not consider the actual
meaning of any constituent identifiers.
[Example: For example, if x and y are identifiers, then x.y is correct
grammar for a type, even if x.y
doesn?t actually denote a type. end example]
[Note: From the disambiguation rule, it follows that, if x and y are
identifiers, (x)y, (x)(y), and (x)(-y)
are cast-expressions, but (x)-y is not, even if x identifies a type.
However, if x is a keyword that identifies
a predefined type (such as int), then all four forms are cast-expressions
(because such a keyword could not
possibly be an expression by itself). end note]
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