我如何可以将三元运营商到precedence爬山算法？ [英] How can I incorporate ternary operators into a precedence climbing algorithm?

问题描述

我跟着precedence爬坡章节中关于此网页的说明的实现使用各种一元preFIX和二进制中缀运算符precedence登山算法算术评估。我也想包括三元运营商（即三元条件运算符：）。

I followed the explanation given in the "Precedence climbing" section on this webpage to implement an arithmetic evaluator using the precedence climbing algorithm with various unary prefix and binary infix operators. I would also like to include ternary operators (namely the ternary conditional operator ?:).

在网页上给出的算法使用以下语法：

The algorithm given on the webpage uses the following grammar:

E --> Exp(0) 
Exp(p) --> P {B Exp(q)} 
P --> U Exp(q) | "(" E ")" | v
B --> "+" | "-"  | "*" |"/" | "^" | "||" | "&&" | "="
U --> "-"

我如何可以将三元运营商进入这个语法？

How can I incorporate ternary operators into this grammar?

推荐答案

要具体，我将使用C / C ++ / Java的：作为一个例子

To be specific, I'll use C/C++/Java's ?: as an example.

目前看来，在这些语言的：运营商允许 之间的任何有效的前pression和？ ：，包括形成前pressions：本身和那些与运营商，他们的precedence形成比的的precedence低：，例如： = 和，（例如： A B = C：D ， A b，C：D ， A b C：？？d：E ）

It appears that in those languages the ?: operator allows any valid expression between ? and :, including expressions formed with ?: itself and those formed with operators, whose precedence is lower than the precedence of ?:, e.g. = and , (examples: a ? b = c : d, a ? b , c : d, a ? b ? c : d : e).

这意味着你应该把 和：以完全相同的方式 （和）在分析前pressions。当你分析出？ EXPR：，它作为一个整体，是一个二元运算。所以，你解析（表达式）和？ EXPR：以同样的方式，但前者是前pression而后者则是一个二元运算符（内嵌有一个前pression）

This suggests that you should treat ? and : in exactly the same manner as ( and ) while parsing expressions. When you have parsed out ? expr :, it, as a whole, is a binary operator. So, you parse ( expr ) and ? expr : in the same way, but the former is an expression while the latter is a binary operator (with an expression embedded in it).

现在？ EXPR：是一个二元运算符（ A -expr-：2 B 比无异A * B 在binaryness条款），你应该能够支持它pretty很像你已经支持任何其他二元运算符。

Now that ? expr : is a binary operator (a ?-expr-: b being no different than a * b in terms of "binaryness"), you should be able to support it pretty much like any other binary operator that you already support.

个人而言，我不劳烦分裂：到自己的独立的二进制运算。最后，它仍然是一个三元运算符，它必须被链接到3操作数和被视为前pression评估过程中一个整体。如果你正在跟踪你在问题中提到的文章中的方法，并正在创建一个AST，然后你去那里，：有一个左子树，右子节点（如任何其他二进制运算符），此外，一中间子节点

Personally, I'd not take the trouble to split ?: into separate binary operators of their own. In the end, it's still a ternary operator and it must be linked to 3 operands and considered as a whole during expression evaluation. If you are following the approach in the article that you mentioned in the question and are creating an AST, then there you go, ?: has a left child node, a right child node (as any other binary operator) and, additionally, a middle child node.

的precedence - expr的 - ：作为一个整体应该是低的。在C / C ++（和Java中？）这不是最低的，但。它是由你来决定你希望它是什么。

The precedence of ?-expr-: as a whole should be a low one. In C/C++ (and in Java?) it's not the lowest, though. It's up to you to decide what you want it to be.

到目前为止，我们还没有涉及的关联性：。在C / C ++和Java， - EXPR - ：是右结合就像赋值操作符 = 。再次，这是由你来让左结合或保持右结合的。

So far we haven't covered the associativity of ?:. In C/C++ and Java, ?-expr-: is right-associative just like the assignment operator =. Again, it's up to you to make it left-associative or keep it right-associative.

和这样的：

E --> P {B P}
P --> v | "(" E ")" | U P
B --> "+" | "-" | "*" | "/" | "^"
U --> "-"

应该成为这样的：

should become something like this:

E --> P {B P}
P --> v | "(" E ")" | U P
B --> "+" | "-" | "*" | "/" | "^" | "?" E ":"
U --> "-"

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