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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.
Another way to create a function object in C++ is to define a non-explicit conversion function to a function pointer type, a function reference type, or a reference to function pointer type. Assuming the conversion does not discard cv-qualifiers , this allows an object of that type to be used as a function with the same signature as the type it ...
Instances of std::function can store, copy, and invoke any callable target—functions, lambda expressions (expressions defining anonymous functions), bind expressions (instances of function adapters that transform functions to other functions of smaller arity by providing values for some of the arguments), or other function objects.
A function pointer, also called a subroutine pointer or procedure pointer, is a pointer referencing executable code, rather than data. Dereferencing the function pointer yields the referenced function, which can be invoked and passed arguments just as in a normal function call.
This function requires C++ – would not compile as C. It has the same behavior as the preceding example but passes the actual parameter by reference rather than passing its address. A call such as addTwo(v) does not include an ampersand since the compiler handles passing by reference without syntax in the call.
The specification for pass-by-reference or pass-by-value would be made in the function declaration and/or definition. Parameters appear in procedure definitions; arguments appear in procedure calls. In the function definition f(x) = x*x the variable x is a parameter; in the function call f(2) the value 2 is the argument of the function. Loosely ...
Unlike the regular double-negation translation, which maps atomic propositions p to ((p → ⊥) → ⊥), the continuation passing style replaces ⊥ by the type of the final expression. Accordingly, the result is obtained by passing the identity function as a continuation to the CPS expression, as in the above example.
The downwards funarg problem complicates the efficient compilation of tail calls and code written in continuation-passing style. In these special cases, the intent of the programmer is (usually) that the function run in limited stack space, so the "faster" behavior may actually be undesirable. [clarification needed]