OF EXERCISES

How to Think like Shakespeare Pub Date : 2020-04-21 DOI:10.2307/j.ctvr7f6m7.12

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Abstract

([ ], [ ]) var z = 2; −−−−−−−→Def1 ([(z, rz)], [(rz, 2)]) let x = 8; −−−−−−−→Def1 ([(x, 8), (z, rz)], [(rz, 2)]) var y = 5; −−−−−−−→Def1 ([(y, ry), (x, 8), (z, rz)], [(ry, 5), (rz, 2)]) let x = 3; −−−−−−−→Def1 etat0 There are several sequences of definitions, which lead to state etat0. For example, in the sequence d, the definition let x = 8 may be replaced by let x =!z + 6 since, in state ([(z, rz)], [(rz, 2)]), the expression !z + 6 is evaluated as 8.

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([]， []) var z = 2;−−−−−−→Def1 ([(z, rz)]， [(rz, 2)]) let x = 8;−−−−−−−→Def1 (((x, 8)、(z, rz)], [(rz, 2)]) var y = 5;−−−−−−−→Def1 (((y, ry), (x, 8)、(z, rz)], [(, 5), (rz, 2)])让x = 3;−−−−−−→Def1 etat0有几种定义顺序，导致状态etat0。例如，在序列d中，定义let x = 8可以替换为let x =!因为，在状态([(Z, rz)]， [(rz, 2)])下，表达式Z + 6的求值为8。

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How to Think like Shakespeare

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