In the expression $ ( o \; r \; \; ) $, "or" is considered a special form because it evaluates the expressions $ $ and $ $ in a way that allows for short-circuit evaluation, meaning that it returns the first truthy result without necessarily evaluating both expressions, unlike standard procedures which always evaluate all their arguments.

Question

In the expression $ ( o \; r \; <\text{exp r 1}> \; <\text{exp r 2}>) $, "or" is considered a special form because it evaluates the expressions $ <\text{exp r 1}> $ and $ <\text{exp r 2}> $ in a way that allows for short-circuit evaluation, meaning that it returns the first truthy result without necessarily evaluating both expressions, unlike standard procedures which always evaluate all their arguments.

Knowee AI · Accepted Answer

In the expression $ ( o \; r \; <\text{exp r 1}> \; <\text{exp r 2}>) $, "or" is considered a special form because it evaluates the expressions $ <\text{exp r 1}> $ and $ <\text{exp r 2}> $ in a way that allows for short-circuit evaluation, meaning that it returns the first truthy result without necessarily evaluating both expressions, unlike standard procedures which always evaluate all their arguments.

Note that in the expression( o r <exp r 1> <exp r 2>)or is considered to be a special form and not a procedure, Why?(Answer in one sentence)

Question

Note that in the expression `( o r <exp r 1> <exp r 2>)`, `or` is considered to be a special form and not a procedure, Why?

Solution

Similar Questions

Upgrade your grade with Knowee