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  ANSI Common Lisp   3 Evaluation and Compilation   3.1 Evaluation

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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