本文翻譯自：Is C++ context-free or context-sensitive?
I often hear claims that C++ is a context-sensitive language. 我經常聽到有人聲稱C ++是上下文相關的語言。 Take the following example: 請看以下示例：
a b(c);
Is this a variable definition or a function declaration? 這是變量定義還是函數聲明？ That depends on the meaning of the symbol c . 這取決於符號c的含義。 If c is a variable , then ab(c); 如果c是一個變量 ，則ab(c); defines a variable named b of type a . 定義了一個名爲變量b類型的a 。 It is directly initialized with c . 它直接用c初始化。 But if c is a type , then ab(c); 但是如果c是一個type ，則ab(c); declares a function named b that takes a c and returns an a . 聲明一個名爲b的函數，該函數採用c並返回a 。
If you look up the definition of context-free languages, it will basically tell you that all grammar rules must have left-hand sides that consist of exactly one non-terminal symbol. 如果您查找無上下文語言的定義，它將基本上告訴您所有語法規則的左手邊必須由一個非終結符組成。 Context-sensitive grammars, on the other hand, allow arbitrary strings of terminal and non-terminal symbols on the left-hand side. 另一方面，上下文相關的語法允許在左側使用任意字符串的終止符和非終止符。
Browsing through Appendix A of "The C++ Programming Language", I couldn't find a single grammar rule that had anything else besides a single non-terminal symbol on its left-hand side. 瀏覽“ C ++編程語言”的附錄A，我找不到一個語法規則，該語法規則的左側除了單個非終止符號外還具有其他內容。 That would imply that C++ is context-free. 這意味着C ++是無上下文的。 (Of course, every context-free language is also context-sensitive in the sense that the context-free languages form a subset of the context-sensitive languages, but that is not the point.) （當然，每種上下文無關的語言也是上下文敏感的，因爲上下文無關的語言構成了上下文敏感語言的子集，但這不是重點。）
So, is C++ context-free or context-sensitive? 那麼，C ++是上下文無關的還是上下文相關的？
#1樓
參考：https://stackoom.com/question/zDM2/C-是上下文無關的還是上下文相關的
#2樓
True :) 是的:)
J. Stanley Warford. 斯坦利·沃福德（J. Stanley Warford）。 Computer systems . 計算機系統 。 Pages 341-346. 341-346頁。
#3樓
C++ is parsed with GLR parser. C ++使用GLR解析器進行解析。 That means during parsing the source code, the parser may encounter ambiguity but it should continue and decide which grammar rule to use later . 這意味着在解析源代碼期間，解析器可能會遇到歧義，但它應該繼續並決定以後使用哪種語法規則。
look also, 也看
Why C++ cannot be parsed with a LR(1) parser? 爲什麼C ++無法使用LR（1）解析器進行解析？
Remember that context-free grammar can not describe ALL the rules of a programming language syntax. 請記住，上下文無關的語法不能描述編程語言語法的所有規則。 For example, Attribute grammar is used to check the validity of an expression type. 例如，屬性語法用於檢查表達式類型的有效性。
int x;
x = 9 + 1.0;
You can not describe the following rule with context-free grammar : The Right Side of the assignment should be of the same type of the Left Hand side. 您不能使用上下文無關的語法描述以下規則： 分配的右側應與左側相同。
#4樓
C++ is not context free. C ++不是上下文無關的。 I learned it some time ago in compilers lecture. 我前一段時間在編譯器講座中學到的。 A quick search gave this link, where the "Syntax or semantics" section explains why C and C++ are not context free: 快速搜索給出了此鏈接，其中“語法或語義”部分說明了爲什麼C和C ++不是上下文無關的：
Wikipedia Talk: Context-Free grammar 維基百科談話：無上下文語法
Regards, 問候，
Ovanes 奧萬斯
#5樓
Yes. 是。 The following expression has a different order of operations depending on type resolved context : 以下表達式根據類型解析的上下文具有不同的操作順序 ：
Edit: When the actual order of operation varies, it makes it incredibly difficult to use a "regular" compiler that parses to an undecorated AST before decorating it (propagating type information). 編輯：當實際的操作順序變化時，將很難使用一個“常規”編譯器來解析未修飾的AST，然後再對其進行修飾（傳播類型信息）。 Other context sensitive things mentioned are "rather easy" compared to this (not that template evaluation is at all easy). 與此相比，提到的其他上下文相關的東西“相當容易”（並不是說模板評估一點都不容易）。
#if FIRST_MEANING
template<bool B>
class foo
{ };
#else
static const int foo = 0;
static const int bar = 15;
#endif
Followed by: 其次是：
static int foobar( foo < 2 ? 1 < 1 : 0 > & bar );
#6樓
有時情況更糟： 當人們說C ++具有“不確定的語法”時，這意味着什麼？