git @ Cat's Eye Technologies Burro / 73a67b4
Continue to work out the extensible conditional idiom. Chris Pressey 4 years ago
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215215 The next step will be combining these conditionals so that we can build a test
216216 that tests for multiple cases.
217217
218 We note that we only have conditionals for greater than and less then.
219 We don't have a conditional test for equality. We can work around that.
220
221 We also note that we can't nest conditionals, because the original value
222 is not available inside the conditional -- it's "hiding" on the stack
223 during that period. We can work around that too.
218 We've already noted that the original value being tested against isn't available
219 inside the conditional -- it's "hiding" on the stack during that period.
220
221 This rules out being able to test multiple cases by nesting conditionals.
222 So instead of nesting conditionals, we'll chain them one after another.
223
224 But there are some implications for this, when it's combined with the fact
225 that we don't have conditional tests for equality, only tests for greater than
226 and less then.
227
228 We'll revert to a more conventional syntax to try to devise how we can overcome
229 those implications.
230
231 Say the value to be tested is either 1, 3, or 5, and say we want to execute
232 _a_ if it's 1, _b_ if it's 3, or _c_ if it's 5. We could try to write our chain
233 like this:
234
235 if value > 0:
236 A
237 endif
238 if value > 2:
239 B
240 endif
241 if value > 4:
242 C
243 endif
244
245 The problem is that A, B, and C don't match up exactly to _a_, _b_, and _c_.
246 Instead we have:
247
248 * If value is 1 then A is executed.
249 * If value is 3 then A is executed then B is executed.
250 * If value is 5 then A then B then C is executed.
251
252 But we can overcome this by defining what A, B, and C are, in terms of
253 _a_, _b_, and _c_, and code that **undoes** _a_, _b_, and _c_. Using
254 the notation _x_′ to mean code that undoes _x_, we can choose the following
255 equations:
256
257 * A = _a_
258 * B = _a_′ _b_
259 * C = _b_′ _c_
260
261 And we can restate the scenario like
262
263 * If value is 1 then _a_ is executed.
264 * If value is 3 then _a_ is executed then _a_′ _b_ is executed.
265 * _a_ _a_′ _b_ = _b_, so in effect, only _b_ is executed.
266 * If value is 5 then _a_ then _a_′ _b_ then _b_′ _c_ is executed.
267 * _a_ _a_′ _b_ _b_′ _c_ = _c_, so in effect, only _c_ is executed.
268
269 So the idiom works out (in theory: now I need to write some test cases here.)