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In the untyped lambda calculus, where the basic types are functions, lifting may change the result of beta reduction of a lambda expression. The resulting functions will have the same meaning, in a mathematical sense, but are not regarded as the same function in the untyped lambda calculus. See also intensional versus extensional equality.
As of Java 8, Java supports functions as first class objects. Lambda expressions of this form are considered of type Function<T,U> with T being the domain and U the image type. The expression can be called with its .apply(T t) method, but not with a standard method call.
Higher-order programming is a style of computer programming that uses software components, like functions, modules or objects, as values. It is usually instantiated with, or borrowed from, models of computation such as lambda calculus which make heavy use of higher-order functions. A programming language can be considered higher-order if ...
In this Erlang example, the higher-order function or_else/2 takes a list of functions (Fs) and argument (X). It evaluates the function F with the argument X as argument. If the function F returns false then the next function in Fs will be evaluated. If the function F returns {false, Y} then the next function in Fs with argument Y will be evaluated.
Anonymous functions are often arguments being passed to higher-order functions or used for constructing the result of a higher-order function that needs to return a function. [1] If the function is only used once, or a limited number of times, an anonymous function may be syntactically lighter than using a named function.
Meta-functions will be given that describe the conversion between lambda and let expressions. A meta-function is a function that takes a program as a parameter. The program is data for the meta-program. The program and the meta program are at different meta-levels. The following conventions will be used to distinguish program from the meta program,
If you’re stuck on today’s Wordle answer, we’re here to help—but beware of spoilers for Wordle 1258 ahead. Let's start with a few hints.
Calling f with a regular function argument first applies this function to the value 2, then returns 3. However, when f is passed to call/cc (as in the last line of the example), applying the parameter (the continuation) to 2 forces execution of the program to jump to the point where call/cc was called, and causes call/cc to return the value 2.