野牛:如何解决减少/减少冲突 [英] Bison: how to fix reduce/reduce conflict

问题描述

下面是一个Bison语法，它说明了我的问题.我正在使用的实际语法更加复杂.

Below is a a Bison grammar which illustrates my problem. The actual grammar that I'm using is more complicated.

%glr-parser
%%
s : e | p '=' s;
p : fp | p ',' fp;
fp : 'x';
e : te | e ';' te;
te : fe | te ',' fe;
fe : 'x';

一些输入示例为:

x
x = x
x,x = x,x
x,x = x;x
x,x,x = x,x;x,x
x = x,x = x;x

我要寻找的是将'='左侧的x解析为与右侧的x不同.但是，可能在"="符号右侧出现的一组合法表达式"要比在左侧的合法表达式"大(因为有;").

What I'm after is for the x's on the left side of an '=' to be parsed differently than those on the right. However, the set of legal "expressions" which may appear on the right of an '='-sign is larger than those on the left (because of the ';').

Bison打印消息(输入文件为test.y):

Bison prints the message (input file was test.y):

test.y: conflicts: 1 reduce/reduce.

必须有一些解决此问题的方法.在C中，您有类似的情况.下面的程序顺利通过了gcc.

There must be some way around this problem. In C, you have a similar situation. The program below passes through gcc with no errors.

int main(void) {
  int x;
  int *px;
  x;
  *px;
  *px = x = 1;
} 

在这种情况下，'px'和'x'的处理方式有所不同，具体取决于它们是出现在'='符号的左侧还是右侧.

In this case, the 'px' and 'x' get treated differently depending on whether they appear to the left or right of an '='-sign.

推荐答案

您正在使用%glr-parser，因此无需修复"减少/减少冲突.野牛只告诉您一个，所以您知道语法可能是模棱两可的，因此您可能需要使用%dprec或%merge指令添加歧义度.但是，对于您而言，语法不是模棱两可的，因此您无需执行任何操作.

You're using %glr-parser, so there's no need to "fix" the reduce/reduce conflict. Bison just tells you there is one, so that you know you grammar might be ambiguous, so you might need to add ambiguity resolution with %dprec or %merge directives. But in your case, the grammar is not ambiguous, so you don't need to do anything.

冲突不是错误，仅表示您的语法不是LALR(1).

A conflict is NOT an error, its just an indication that your grammar is not LALR(1).

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