C ++编译器如何区分令牌>>对于二进制运算符，对于模板 [英] How C++ compilers differentiate the token >> for binary operator, and for template

问题描述

我的疑惑是关于C ++编译器作为Clang的解析器，编译器如何处理运算符>> 以了解其何时是二进制运算符以及何时关闭模板，如: std :: vector< std :: tuple< int，double>> ，我想这是在解析器时间内完成的，因此解决该问题的更好方法是在词法上或仅使用>作为标记，并且解决语法解析器中的问题?

My doubt is about the parser of C++ compilers as Clang, how the compilers handle the operator >> to know when it is a binary operator and when it is closing a template like: std::vector<std::tuple<int, double>>, I imagine that is done in parser time, so the better way to solve that is on lexical or use only > as token, and solve the problem in the grammar parser?

推荐答案

实际上很简单:如果可见打开的模板支架，即使> 否则将构成>> 运算符的一部分.(这不适用于作为其他标记一部分的> 字符，例如> = .)对C ++语法的此更改是C ++ 11的一部分，并在§13.3[temp.names]的第3段中进行了描述.

It's actually quite simple: if there is an open template bracket visible, a > closes it, even if the > would otherwise form part of a >> operator. (This doesn't apply to > characters which are part of other tokens, such as >=.) This change to C++ syntax was part of C++11, and is described in paragraph 3 of §13.3 [temp.names].

如果> 在带括号的嵌套语法内，则打开的模板括号不可见.因此，两个 T< sizeof a [x>>中的>> 1]> 和 T<(x>> 1)> 是右移运算符，而 T< x>1> 可能无法按预期进行解析.

An open template bracket is not visible if the > is inside a parenthetically nested syntax. So the >> in both T<sizeof a[x >> 1]> and T<(x >> 1)> are right shift operators, while T<x >> 1> probably does not parse as expected.

这两种实现策略都是可行的，具体取决于您要放置复杂性的位置.如果词法分析器从不生成>> 令牌；解析器可以检查 expr'>'中的> 标记'>'expr 通过查看其源位置而相邻.将存在移位减少冲突，必须解决该冲突以减少模板参数列表.之所以行之有效是因为碰巧不会通过将>> 分成两个标记而产生歧义，但这不是一般规则: a + ++ b 与 a ++ + b ;如果词法分析器仅生成 + 令牌，那将是模棱两可的.

The two implementation strategies are both workable, depending on where you want to put the complexity. If the lexer never generates a >> token; the parser can check that the > tokens in expr '>' '>' expr are adjacent by looking at their source locations. There will be a shift-reduce conflict, which will have to be resolved in favour of reducing the template parameter list. This works because it happens that there is no ambiguity created by separating >> into two tokens, but that's not a general rule: a + ++ b is different from a ++ + b; if the lexer were only generating + tokens, that would be ambiguous.

如果您准备好使用词法分析器跟踪括号深度，使用词法分析器hack解决该问题并不复杂.这意味着词法分析器必须知道< 是模板括号还是比较运算符，但很有可能确实如此.

It's not too complicated to resolve the issue with a lexer hack, if you are prepared to have your lexer track parenthesis depth. That means the lexer has to know whether a < is a template bracket or a comparison operator, but it is quite possible that it does.

这是一个更有趣的问题(至少是恕我直言):如何将< 识别为模板括号而不是小于运算符?这里确实有语义反馈:如果它跟随指定模板的名称，则它是模板支架.

This is the more interesting question (at least imho): how is a < recognised as a template bracket rather than a less-than operator? Here there really is semantic feedback: it is a template bracket if it follows a name which designates a template.

这不是一个简单的决定.该名称可以是类或联合的成员，甚至可以是模板化类或联合的专业化的成员.在后一种情况下，可能有必要计算编译时常量表达式的值，然后进行模板推导，以确定名称的含义.

This is not a simple determination. The name could be a class or union member, and even a member of a specialisation of a templated class or union. In the latter case, it might be necessary to calculate the values of compile-time constant expressions and then do template deduction in order to decide what the name designates.

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