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In contrast to well-formed nested parentheses and square brackets in the previous section, there is no context-free grammar for generating all sequences of two different types of parentheses, each separately balanced disregarding the other, where the two types need not nest inside one another, for example: [ ( ] ) or
The most vexing parse is a counterintuitive form of syntactic ambiguity resolution in the C++ programming language. In certain situations, the C++ grammar cannot distinguish between the creation of an object parameter and specification of a function's type.
Starting after the second symbol, match the shortest subexpression y of x that has balanced parentheses. If x is a formula, there is exactly one symbol left after this expression, this symbol is a closing parenthesis, and y itself is a formula. This idea can be used to generate a recursive descent parser for formulas. Example of parenthesis ...
The detailed semantics of "the" ternary operator as well as its syntax differs significantly from language to language. A top level distinction from one language to another is whether the expressions permit side effects (as in most procedural languages) and whether the language provides short-circuit evaluation semantics, whereby only the selected expression is evaluated (most standard ...
An example where this does not work is the logical biconditional ↔. It is associative; thus, A ↔ (B ↔ C) is equivalent to (A ↔ B) ↔ C, but A ↔ B ↔ C most commonly means (A ↔ B) and (B ↔ C), which is not equivalent.
(e.g. (a * b) * c = a * (b * c)). Many programming language manuals provide a table of operator precedence and associativity; see, for example, the table for C and C++ . The concept of notational associativity described here is related to, but different from, the mathematical associativity .
In programming languages where assignment is implemented as an operator, that operator is often right-associative. If so, a statement like a := b := c would be equivalent to a := (b := c), which means that the value of c is copied to b which is then copied to a. An operator which is non-associative cannot compete for operands with operators of ...
The proof that the language of balanced (i.e., properly nested) parentheses is not regular follows the same idea. Given p {\displaystyle p} , there is a string of balanced parentheses that begins with more than p {\displaystyle p} left parentheses, so that y {\displaystyle y} will consist entirely of left parentheses.