Extent (ANSI and GNU Common Lisp Document)

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#### 3.1.6 Extent

`Contorted-example` works only because the function named by `f` is invoked during the extent of the exit point. Once the flow of execution has left the block, the exit point is disestablished. For example:

     (defun invalid-example ()
       (let ((y (block here #'(lambda (z) (return-from here z)))))
         (if (numberp y) y (funcall y 5))))

One might expect the call `(invalid-example)` to produce `5` by the following incorrect reasoning: let binds `y` to the value of block; this value is a function resulting from the lambda expression. Because `y` is not a number, it is invoked on the value `5`. The return-from should then return this value from the exit point named `here`, thereby exiting from the block again and giving `y` the value `5` which, being a number, is then returned as the value of the call to `invalid-example`.

The argument fails only because exit points have dynamic extent. The argument is correct up to the execution of return-from. The execution of return-from should signal an error of type control-error, however, not because it cannot refer to the exit point, but because it does correctly refer to an exit point and that exit point has been disestablished.

A reference by name to a dynamic exit point binding such as a catch tag refers to the most recently established binding of that name that has not been disestablished. For example:

     (defun fun1 (x)
       (catch 'trap (+ 3 (fun2 x))))
     (defun fun2 (y)
       (catch 'trap (* 5 (fun3 y))))
     (defun fun3 (z)
       (throw 'trap z))

Consider the call `(fun1 7)`. The result is `10`. At the time the throw is executed, there are two outstanding catchers with the name `trap`: one established within procedure `fun1`, and the other within procedure `fun2`. The latter is the more recent, and so the value `7` is returned from catch in `fun2`. Viewed from within `fun3`, the catch in `fun2` shadows the one in `fun1`. Had `fun2` been defined as

     (defun fun2 (y)
       (catch 'snare (* 5 (fun3 y))))

then the two exit points would have different names, and therefore the one in `fun1` would not be shadowed. The result would then have been `7`.

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