我经常听到有人说c++是一种上下文敏感的语言。举个例子:
a b(c);
这是变量定义还是函数声明?这取决于符号c的含义。如果c是一个变量,则ab (c);定义一个名为b的类型为a的变量。它直接用c初始化。但如果c是类型,则ab (c);声明一个名为b的函数,该函数接受c并返回a。
如果您查找上下文无关语言的定义,它基本上会告诉您,所有语法规则的左侧必须恰好包含一个非终结符。另一方面,上下文敏感语法允许在左侧使用任意的终结符和非终结符字符串。
浏览“c++程序设计语言”的附录A,我找不到一条语法规则,它的左边除了一个非终结符之外,还有其他任何东西。这意味着c++是上下文无关的。(当然,每一种与上下文无关的语言也是与上下文相关的,因为与上下文无关的语言构成了与上下文相关的语言的一个子集,但这不是重点。)
那么,c++是上下文无关的还是上下文敏感的?
这里的一个大问题是术语“上下文无关”和“上下文敏感”在计算机科学中有点不直观。对于c++,上下文敏感性看起来很像歧义,但在一般情况下不一定是这样。
在C/ c++中,if语句只允许在函数体中使用。这似乎是上下文敏感的,对吧?嗯,没有。与上下文无关的语法实际上不需要这样的属性,即您可以提取某一行代码并确定它是否有效。这实际上并不是context-free的意思。它实际上只是一个标签,模糊地暗示了一些与它的发音相关的东西。
现在,如果函数体中的语句根据直接语法祖先之外的定义而被不同地解析(例如,标识符是描述类型还是变量),如a * b;大小写,那么它实际上是上下文敏感的。这里没有实际的歧义;如果a是类型,则解析为指针声明,否则解析为乘法声明。
Being context-sensitive does not necessarily mean "hard to parse". C is actually not that hard because the infamous a * b; "ambiguity" can be resolved with a symbol table containing typedefs encountered previously. It doesn't require any arbitrary template instantiations (which have been proven to be Turing Complete) to resolve that case like C++ does on occasion. It's not actually possible to write a C program that will not compile in a finite amount of time even though it has the same context-sensitivity that C++ does.
Python (and other whitespace-sensitive languages) is also context-dependent, as it requires state in the lexer to generate indent and dedent tokens, but that doesn't make it any harder to parse than a typical LL-1 grammar. It actually uses a parser-generator, which is part of why Python has such uninformative syntax error messages. It's also important to note here that there is no "ambiguity" like the a * b; problem in Python, giving a good concrete example of a context-sensitive language without "ambiguous" grammar (as mentioned in the first paragraph).
这里的一个大问题是术语“上下文无关”和“上下文敏感”在计算机科学中有点不直观。对于c++,上下文敏感性看起来很像歧义,但在一般情况下不一定是这样。
在C/ c++中,if语句只允许在函数体中使用。这似乎是上下文敏感的,对吧?嗯,没有。与上下文无关的语法实际上不需要这样的属性,即您可以提取某一行代码并确定它是否有效。这实际上并不是context-free的意思。它实际上只是一个标签,模糊地暗示了一些与它的发音相关的东西。
现在,如果函数体中的语句根据直接语法祖先之外的定义而被不同地解析(例如,标识符是描述类型还是变量),如a * b;大小写,那么它实际上是上下文敏感的。这里没有实际的歧义;如果a是类型,则解析为指针声明,否则解析为乘法声明。
Being context-sensitive does not necessarily mean "hard to parse". C is actually not that hard because the infamous a * b; "ambiguity" can be resolved with a symbol table containing typedefs encountered previously. It doesn't require any arbitrary template instantiations (which have been proven to be Turing Complete) to resolve that case like C++ does on occasion. It's not actually possible to write a C program that will not compile in a finite amount of time even though it has the same context-sensitivity that C++ does.
Python (and other whitespace-sensitive languages) is also context-dependent, as it requires state in the lexer to generate indent and dedent tokens, but that doesn't make it any harder to parse than a typical LL-1 grammar. It actually uses a parser-generator, which is part of why Python has such uninformative syntax error messages. It's also important to note here that there is no "ambiguity" like the a * b; problem in Python, giving a good concrete example of a context-sensitive language without "ambiguous" grammar (as mentioned in the first paragraph).
是的,c++是上下文敏感的,非常上下文敏感。您不能通过使用上下文无关的解析器简单地解析文件来构建语法树,因为在某些情况下,您需要从以前的知识中了解符号来决定(例如。在解析时构建一个符号表)。
第一个例子:
A*B;
这是一个乘法表达式吗?
OR
这是B变量作为a类型指针的声明吗?
如果A是一个变量,那么它就是一个表达式,如果A是类型,它就是一个指针声明。
第二个例子:
A B(bar);
这是一个函数原型接受一个条形参数吗?
OR
这是声明A类型的变量B,并调用A的构造函数和bar常量作为初始化式吗?
您需要再次了解bar是符号表中的变量还是类型。
第三个例子:
class Foo
{
public:
void fn(){x*y;}
int x, y;
};
当解析时构建符号表没有帮助时,就是这种情况,因为x和y的声明在函数定义之后。因此,您需要首先浏览类定义,然后在第二步查看方法定义,以确定x*y是一个表达式,而不是指针声明或其他东西。