关于符号的定义和值的问题
Definition“知道”符号值的定义方式：使用Set或SetDelayed。但如何呢？据我了解，在为符号分配值后，评估器的分配方式没有任何区别：使用 Set 或 SetDelayed。它可以通过函数 OwnValues 来说明，该函数始终返回带有 Head RuleDelayed 的定义。 Definiton 如何获取此信息？
In[1]:= a=5;b:=5;
Definition[a]
Definition[b]
OwnValues[a]
Out[2]= a=5
Out[3]= b:=5
Out[4]= {HoldPattern[a]:>5}
Definition "knows" the way how a value for a symbol was defined: using Set or SetDelayed. But how? As I understand, after a value for a symbol was assigned there is no any difference for the evaluator how it was assigned: by using Set or SetDelayed. It can be illustrated by the function OwnValues which always returns definitions with the Head RuleDelayed. How Definiton obtains this information?
In[1]:= a=5;b:=5;
Definition[a]
Definition[b]
OwnValues[a]
Out[2]= a=5
Out[3]= b:=5
Out[4]= {HoldPattern[a]:>5}
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OwnValues[a] = {HoldPattern[a] ->; 3}； OwnValues[a]给出{HoldPattern[a] :> 3}而不是{HoldPattern[a] -> 3}但Definition[a]显示了人们可以期待的内容。该定义可能以Rule的形式在内部存储，但通过OwnValues转换为RuleDelayed，以抑制对定义的 rhs 的评估。这个假设与我最初的理解相矛盾，即 Set 和 SetDelayed 分配的值之间没有区别。这些定义可能以不同的形式存储：相应的Rule和RuleDelayed，但从评估者的角度来看是等效的。有趣的是，
MemoryInUse[]如何取决于定义的类型。在下面的实验中，我在没有前端的交互式会话中使用了 Mathematica 5.2 的内核。使用 Mathematica 6 和 7 的内核将会得到不同的结果。原因之一是在这些版本中
Set< /code> 默认情况下重载。首先，我评估
$HistoryLength=0;是否有DownValues的In和Out变量不影响我的结果。但似乎即使当$HistoryLength设置为 0 时，当前 输入行的In[$Line]值仍然会被存储和删除输入新的输入后。这可能就是为什么MemoryInUse[]的第一次评估结果总是与第二次不同的原因。这是我得到的：
可以看到，定义
a=2;将MemoryInUse[]增加1986824-1986760=64 字节。将其替换为定义a:=2;会使MemoryInUse[]增加 1986952-1986824=128 字节。将后一个定义替换为前一个定义会将MemoryInUse[]恢复为 1986824 字节。这意味着延迟定义比立即定义多需要 128 个字节。当然这个实验并不能证明我的假设。
OwnValues[a] = {HoldPattern[a] -> 3}; OwnValues[a]gives{HoldPattern[a] :> 3}instead of{HoldPattern[a] -> 3}butDefinition[a]shows what one can expect. Probably this definition is stored internally in the form ofRulebut is converted toRuleDelayedbyOwnValuesfor suppressing of evaluation of the r.h.s of the definition. This hypothesis contradicts my original understanding that there are no difference between values assigned bySetandSetDelayed. Probably such definitions are stored in different forms:RuleandRuleDelayedcorrespondingly but are equivalent from the evaluator's point of view.It is interesting to see how
MemoryInUse[]depends on the kind of definition.In the following experiment I used the kernel of Mathematica 5.2 in interactive session without the FrontEnd. With the kernels of Mathematica 6 and 7 one will get different results. One reason for this is that in these versions
Setis overloaded by default.First of all I evaluate
$HistoryLength=0;for havingDownValuesforInandOutvariables not affecting my results. But it seems that even when$HistoryLengthis set to 0 the value ofIn[$Line]for current input line is still stored and removed after entering new input. This is likely the reason why result of the first evaluation ofMemoryInUse[]always differs from the second.Here is what I have got:
One can see that defining
a=2;increasesMemoryInUse[]by 1986824-1986760=64 bytes. Replacing it with the definitiona:=2;increasesMemoryInUse[]by 1986952-1986824=128 bytes. And replacing the latter definition with the former revertsMemoryInUse[]to 1986824 bytes. It means that delayed definitions require 128 bytes more than immediate definitions.Of course this experiment does not prove my hypothesis.
可以通过未记录的 new-in-8 符号
Language`ExtendedDefinition和Language`ExtendedFullDefinition访问符号的完整定义。引用Oleksandr Rasputinov：“如果有人很好奇，
Language`ExtendedDefinition和Language`ExtendedFullDefinition是与Definition和FullDefinition类似，但以可以在另一个内核中重现的方式捕获符号的定义，例如，defs = Language`ExtendedFullDefinition。 [sym]返回一个Language`DefinitionList对象 用于恢复定义的语法非常不规则：Language`ExtendedFullDefinition[] = defs，其中。defs是Language`DefinitionList请注意，Language`ExtendedFullDefinition采用ExcludedContexts选项，而Language` ExtendedDefinition没有。”Complete definition for a symbol can be accessed via undocumented new-in-8 symbols
Language`ExtendedDefinitionandLanguage`ExtendedFullDefinition. Citing Oleksandr Rasputinov:"If anyone is curious,
Language`ExtendedDefinitionandLanguage`ExtendedFullDefinitionare analogous toDefinitionandFullDefinitionbut capture the definition of a symbol in such a way as it can be reproduced in another kernel. For example,defs = Language`ExtendedFullDefinition[sym]returns aLanguage`DefinitionListobject. The syntax used to restore the definition is highly irregular:Language`ExtendedFullDefinition[] = defs, wheredefsis aLanguage`DefinitionList. Note thatLanguage`ExtendedFullDefinitiontakes theExcludedContextsoption whereasLanguage`ExtendedDefinitiondoes not."Information调用Definition，并且Definition（或FullDefinition）上的跟踪未显示任何内容。我必须假设这是一个访问*Values表之外的数据的低级函数。也许它保留了当时解析的原始定义表达式的副本。InformationcallsDefinition, and a Trace onDefinition(orFullDefinition) shows nothing. I must assume that this is a low level function that accesses data outside of the*Valuestables. Perhaps it keeps a copy of the original definition expressions as they were parsed at that time.