When would I use mapc instead of mapcar?

Question

So far I have been using `mapcar` to apply a function to all elements of a list, such as: ``` (mapcar (lambda (x) (* x x)) '(1 2 3 4 5)) ;; => '(1 4 9 16 25) ``` Now I learned that there is also the `mapc` function which does exactly the same, but does not return a new list, but the original one: ``` (mapc (lambda (x) (* x x)) '(1 2 3 4 5)) ;; => '(1 2 3 4 5) ``` What's the intent of this function? When would I use `mapc` instead of `mapcar` if I am not able to access the result?

Accepted Answer

The Common Lisp Hyperspec says:

`mapc` is like `mapcar` except that the results of applying function are not accumulated. The list argument is returned.

So it is used when mapping is done for possible side-effects. `mapcar` could be used, but `mapc` reduces unnecessary consing. Also its return value is the original list, which could be used as input to another function.

Example:

```
(mapc #'delete-file (mapc #'compile-file '("foo.lisp" "bar.lisp")))
```

Above would first compile the source files and then delete the source files. Thus the compiled files would remain.

```
(mapc #'delete-file (mapcar #'compile-file '("foo.lisp" "bar.lisp")))
```

Above would first compile the source files and then delete the compiled files.

When would I use mapc instead of mapcar?

Solution

Problem