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Talk:Function composition: Difference between revisions
Go for it in C
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:::Thats right, it is another aspect of first class functions but there is no need to show functions as members of other collection types. Some languages may be able to do this and not First Class Functions. --[[User:Paddy3118|Paddy3118]] 15:48, 4 March 2009 (UTC)
::Let me show another doubt of mine. What a function ''is'' for a language, is definible inside the same language; C can deal naturally with pointers, and we pass ''function'' to e.g. other functions by pointer (reference?); so foo(bar) call foo with argument bar (by the way, "calling a subroutine" for compiled languages means always to know its address, even at the end of the games a run-time, such the one of Objective-C when "finding" the code for a selector, ''compute'' the address of other compiled code...), and "foo" ''is'' the function (which C handles by pointer). So this task, if the right constraint of [[First-class functions]] is dropped (exactly the first: "New functions can be created from others at run time"), can be implemented in C. If it is not droppable, languages can accomplish this task iff they "have" first class functions. Am I reasonably right? --[[User:ShinTakezou|ShinTakezou]] 16:23, 4 March 2009 (UTC)
::IF you can create a version of compose that works for function pointers when you have an arbitrary set of two functions to compose then why not put it down? I add the extra restriction because if you had n sets of f and g functions to compose, i.e.
composeN = compose(fN, gN) for N in 0..N-1
::Then any of the composeN should be able to be used after all have been created. I can think of a naive implementation using function pointers where the last call to compose would reset the actions of all the composeN - this is not what is meant. My C is a little rusty but I guess if you could do something like:
<lang c> fg = compose(&f, &g)
(*fg)(x) /* Where (*fg)(x) == f(g(x)) and another call to compose would leave this one alone */</lang>
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