内容简介:Most programming languages’ grammars fall into the category of Context-Free Grammar (CFG), or sometimes Context-Sensitive Grammar (CSG). What about C++? It turns out parsing C++’s grammar isTo prove this, we will write a program that is parsed differently
Most programming languages’ grammars fall into the category of Context-Free Grammar (CFG), or sometimes Context-Sensitive Grammar (CSG). What about C++? It turns out parsing C++’s grammar is literally undecidable . In other words, if you have a compiler that parses all valid C++ sources perfectly, you have solved the Halting Problem.
To prove this, we will write a program that is parsed differently depending on the solution to the Halting Problem. But we need to implement a Turing Machine to formulate the Halting Problem, and doing that with only template metaprogramming is a huge pain, so we will instead use the Post Correspondence Problem which is proven to be equivalent to the Halting Problem.
The Post Correspondence Problem’s statement is simple: You are given a set of dominoes which have an array of symbols written on the top part and the bottom part. Suppose there are an infinite amount of each type of domino. Can you arrange the dominoes so that the string of symbols on the top part matches the string of symbols on the bottom part?
Turns out, there exists no algorithm that says “yes” or “no” to the Post Correspondence Problem in finite time, given any set of dominoes as input.
We first define a variadic struct template Row , which represents a “row” of symbols ( int s). We add a convenience static member constant that says whether this row is empty.
Now we define a template that represents a domino:
And a handy type alias that concatenates two Row s together:
Now, to exhaustively search through the entire search space, we will use a Breadth-First Search (BFS) algorithm. In the core of BFS lies a FIFO queue, so we first implement that:
Each state of the search space consists of the upper row and lower row of the already-arranged dominoes. We can check if the state is a match by checking if the two rows are identical and not empty.
Now we implement the core of the algorithm. First, pop the head of the queue, and check if the popped state is a match. If it is, the problem is solved, and the answer is “yes”; if it isn’t, push all the child states generated by appending a domino at the right end, and keep going. If the queue is somehow empty, that means we looked at the entire search space and did not find a match, so the answer is “no”.
The initial state is two empty rows:
Now, DominoList::match has the solution to the Post Correspondence Problem defined by the template argument Dominos !
We can now define a helper struct ParseThis whose member typeOrValue is a type int if the solution to the Post Correspondence Problem is “yes”, and a value 0 of type int if the solution is “no”, using SFINAE:
Therefore, the following line is a function declaration if the solution to the Post Correspondence Problem with the dominoes [bba, bb], [ab, aa], [a, baa] is “Yes”, and a variable declaration if it is “No”. (In the former case, it is equivalent to int x(int); , and in the latter case, it is equivalent to int x(0); or int x = 0; )
There we have it! We can substitute any set of dominoes in the above line, and a conforming compiler should be able to decide whether x is a function or variable! If such compiler exists, however, we can use it to solve the Post Correspondence Problem for any input, which means we can use it to solve the Halting Problem. But no such program can exist, which means parsing C++ is undecidable. QED
Read the full code here:
https://gist.github.com/mujjingun/efcdc9d8e82bc44c67843a542d3917d9
以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持 码农网
本站部分资源来源于网络,本站转载出于传递更多信息之目的,版权归原作者或者来源机构所有,如转载稿涉及版权问题,请联系我们。
Web2.0策划指南
艾美 / 2009-11 / 32.00元
《Web2.0策划指南(影印版)》是讲述战略的。书中的示例关注的是Web 20的效率,而不聚焦于技术。你将了解到这样一个事实:创建Web 20业务或将Web 20战略整合到业务中,意味着创建一个吸引人们前来访问的在线站点,让人们愿意到这里来共享他们的思想、见闻和行动。当人们通过Web走到一起时,可能得到总体远远大于各部分和的结果。随着传统的“口碑传诵”助推站点高速成长,客户本身就能够帮助建立站点。......一起来看看 《Web2.0策划指南》 这本书的介绍吧!