C# Language Specification

Cast expressions

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 (§6.2) of the value of E to type T. If no explicit conversion exists from E to T, a binding-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.

The grammar for a cast-expression leads to certain syntactic ambiguities. For example, the expression (x)–y could either be interpreted as a cast-expression (a cast of –y to type x) or as an additive-expression combined with a parenthesized-expression (which computes the value x – y).

To resolve cast-expression ambiguities, the following rule exists: A sequence of one or more tokens (§2.3.3) 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 (§2.4.1), a literal (§2.4.4), or any keyword (§2.4.3) 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. 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.

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).

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