Partial function application: Difference between revisions
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2 4 6 8 ==f2==> 4 16 36 64
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{{header|TXR}}
The <code>f1</code> function is just <code>mapcar</code>, or at least the oft-used special case thereof when there is only one list to process and the function takes one argument.
The <code>op</code> operator in TXR Lisp is used for partial application.
Multiply every element of a list by two, by partially applying <code>2</code> to the <code>*</code> operator:
<lang txr>(mapcar (op * 2) '(1 2 3))</lang>
Now that is not a function, but a function call. If we want a one-argument function in which the <code>mapcar</code> function itself is curried, we use <code>op</code> again:
<lang txr>(op mapcar (op * 2))</lang>
This could be applied to the list like this to produce <code>(2 4 6)</code>:
<lang txr>[(op mapcar (op * 2)) '(1 2 3)]</lang>
We could define names for these functions like the problem statement says, but that defeats much of the point of partial application:
<lang txr>(defun fsf1 (list)
[(op mapcar (op * 2)) '(1 2 3)]) ;; pointless: why don't we just call mapcar?</lang>
Here is an <code>op</code> expressino that returns a function which squares the elements of a sequence, producing a new sequence. Since we use the multiplication function, we must repeat an argument and so we must use the <code>@number</code> syntax supported by <code>op</code>:
<lang txr>(op mapcar (op * @1 @1))</lang>
This is an anonymous version of <code>fsf2</code>.
{{omit from|Euphoria}}
{{omit from|Go|No way to define one function in terms of another.}}
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