关于C和C ++编译器之间的注释解析的问题 [英] Question about comment parsing between C and C++ compiler
问题描述
我正在读书<<专家C编程>>，它有以下
测验，
a // *
// * / b
在C和C ++编译器中，上面的代码是什么？
我认为它对C编译器来说很简单，它是a / b。
但对于C ++编译器，本书说这是一个。原因是//
发表其余的评论。
我想知道这个。
只需回页，它就会提到编译器有一个最大值
munch策略。对于我来说，当C ++编译器读取第一行时，
存在模糊的解释，它可能是a // *。或者a / / *，
然后如果我们应用maximal much strategy，它应该使用第二个
一个并解析代码
a / / *
// * / b
这是a / b。
我觉得我在某个地方很困惑，你能不知所措？
谢谢。
li*****@hotmail.com 写道：
< blockquote class =post_quotes>
>
我正在读书<< Expert C Programming>> ;,它有以下内容
测验，
a // *
// * / b
在C和C ++编译器中，上面的代码中断了什么？
我觉得C编译器很简单，它是a / b。
但是对于C ++编译器，这本书说它是一个。原因是//
发表其余的评论。
我想知道这个。
只需回页，它就会提到编译器有一个最大值
munch策略。对于我来说，当C ++编译器读取第一行时，
存在模糊的解释，它可能是a // *。或者a / / *，
然后如果我们应用maximal much strategy，它应该使用第二个
一个并解析代码
a / / *
// * / b
这是a / b。
我觉得我在某个地方很困惑，你能解开一些光吗？
最大咀嚼策略是什么？是什么意思？
-
pete
li ***** @ hotmail.com 说：
我正在阅读本书<< Expert C Programming>>，它有以下
测验，
a // *
// * / b
在C和C ++编译器中，上面的代码输出了什么？
我觉得它很简单对于C编译器，它是a / b。
这很简单，但现在它并不像你做的那么简单。
out。 PvdL没有意识到的是// - 将在1999年的语言修订版中将评论引入C中给出答案！
但是对于C ++编译器，书中说它是一个。原因是//
进行其余的评论。
是的。
我想知道这个。
只是回到页面，它提到编译器有一个最大的
munch策略。
抓住最大的令牌，是的。
对我来说，当C ++编译器读取第一行时，
存在含糊不清的解释，它可能是a // *或者a / / *，
不，它不可能是其中之一。它可能是
a / / *
或
a // *
和最大的munch决定了第二个。
C ++是否真的有一个最大的蒙克规则是一个问题，我们的
朋友们comp.lang.c ++毫无疑问能够回答。
-
理查德希思菲尔德
Usenet是一个奇怪的放置" - dmr 29/7/1999
http://www.cpax.org.uk
电子邮件：rjh在上述域名中， - www。
pete< pf ***** @ mindspring。 comwrites：
[...]
最大咀嚼策略是什么？意思？
这意味着，在确定下一个令牌时，编译器会尽可能多地抓取许多字符来获取有效令牌。 />
例如，这个：
x +++++ y
被标记为< br $> b $ bx ++ ++ + y
导致语法错误，即使这样：
x ++ + ++ y
将导致有效的解析。 （Tokenization并不代表
后期阶段。）
-
Keith Thompson（The_Other_Keith） ks *** @ mib.org < http://www.ghoti.net/~kst>
San迭戈超级计算机中心< *< http://users.sdsc.edu/~kst>
我们必须做点什么。这是事情。因此，我们必须这样做。
- Antony Jay和Jonathan Lynn，是部长
Hi,
I am reading book <<Expert C Programming>>, it has the following
quiz,
a //*
//*/ b
In C and C++ compiler what does the above code trun out?
I think it is simple for C compiler, it is a/b.
But for C++ compiler, the book says it is a. The reason is "//"
makes the rest of line comment.
I am wondering on this.
Just couple page back, it mentions that compiler has a "maximal
munch strategy". For me when the C++ compiler reads the 1st line,
there is ambiguous intepretation, it could be "a// *" or "a / /*",
then if we apply the "maximal much strategy", it should use the second
one and parse the code to
a / /*
// */ b
which is a/b.
I think I am confused at somewhere, could you shed some light?
Thanks.
li*****@hotmail.com wrote:>
Hi,
I am reading book <<Expert C Programming>>, it has the following
quiz,
a //*
//*/ b
In C and C++ compiler what does the above code trun out?
I think it is simple for C compiler, it is a/b.
But for C++ compiler, the book says it is a. The reason is "//"
makes the rest of line comment.
I am wondering on this.
Just couple page back, it mentions that compiler has a "maximal
munch strategy". For me when the C++ compiler reads the 1st line,
there is ambiguous intepretation, it could be "a// *" or "a / /*",
then if we apply the "maximal much strategy", it should use the second
one and parse the code to
a / /*
// */ b
which is a/b.
I think I am confused at somewhere, could you shed some light?What does "maximal munch strategy" mean?
--
pete
li*****@hotmail.com said:
Hi,
I am reading book <<Expert C Programming>>, it has the following
quiz,
a //*
//*/ b
In C and C++ compiler what does the above code trun out?
I think it is simple for C compiler, it is a/b.It''s fairly simple, but nowadays it is not quite as simple as you make
out. What PvdL didn''t realise was that //-comments would be introduced
into C in the 1999 language revision!
But for C++ compiler, the book says it is a. The reason is "//"
makes the rest of line comment.Yeah.
I am wondering on this.
Just couple page back, it mentions that compiler has a "maximal
munch strategy".Grab the biggest token you can, yes.
For me when the C++ compiler reads the 1st line,there is ambiguous intepretation, it could be "a// *" or "a / /*",No, it can''t be either of those. It could be
a / /*
or
a // *
and maximal munch dictates the second.
Whether C++ actually has a maximal munch rule is a question that our
friends in comp.lang.c++ would undoubtedly be able to answer.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
pete <pf*****@mindspring.comwrites:
[...]What does "maximal munch strategy" mean?It means that, when determining the next token, the compiler grabs as
many characters as possible to get a valid token.
For example, this:
x+++++y
is tokenized as
x ++ ++ + y
which results in a syntax error, even though this:
x ++ + ++ y
would result in a valid parse. (Tokenization doesn''t account for
later phases.)
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
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