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functionclojureanonymous-function

Anonymous function shorthand


There's something I don't understand about anonymous functions using the short notation #(..)

The following works:

REPL>  ((fn [s] s) "Eh")
"Eh"

But this doesn't:

REPL>  (#(%) "Eh")

This works:

REPL> (#(str %) "Eh")
"Eh"

What I don't understand is why (#(%) "Eh") doesn't work and at the same time I don't need to use str in ((fn [s] s) "Eh")

They're both anonymous functions and they both take, here, one parameter. Why does the shorthand notation need a function while the other notation doesn't?


Solution

  • #(...)
    

    is shorthand for

    (fn [arg1 arg2 ...] (...))
    

    (where the number of argN depends on how many %N you have in the body). So when you write:

    #(%)
    

    it's translated to:

    (fn [arg1] (arg1))
    

    Notice that this is different from your first anonymous function, which is like:

    (fn [arg1] arg1)
    

    Your version returns arg1 as a value, the version that comes from expanding the shorthand tries to call it as a function. You get an error because a string is not a valid function.

    Since the shorthand supplies a set of parentheses around the body, it can only be used to execute a single function call or special form.