由预处理器管理的数组 [英] Array managed by preprocessor
问题描述
我的老师说C中的数组是由预处理器管理的。
Preprocesser用
替换所有数组出现（即int [10]）我不理解/记得很清楚。
预处理/编译阶段阵列究竟发生了什么？
谢谢提前
/ * frank * /写道：我的老师说C中的数组是由预处理器管理的。
Preprocesser用一些我不理解/记得不错的东西替换所有阵列出现（即int [10]）。
什么是在预处理/编译阶段确实发生了数组吗？
提前致谢
嗯，这里有两种可能：
1）你误解了老师说的话。
2）你的老师非常可怕，非常错误。
我怀疑你需要修改坐在主题上;为什么不*问你的老师*？
HTH，
-ag
-
Artie Gold - 德克萨斯州奥斯汀
他们指责你 - 是他们的计划。
< br>
Artie Gold ha scritto：
1）你误解了老师说的话。
如果我有这个数组：
a [i]
预处理器将其翻译成*（a + i）
a被翻译为costant。
谢谢
/ * frank * /< __ ******* @ despammed.com>潦草地写道：Artie Gold ha scritto：1）你误解了老师说的话。
如果我有这个数组：
a [i]
预处理器将其翻译成*（a + i）
a翻译为costant。
预处理器没有'不能把它翻译成任何东西。它的工作在
那个阶段完成，它已退出舞台。 []是一个成熟的运算符在
C中。它的行为定义为*（间接）和+（加法）
运算符组合使得a [i]表示*（a + i）。另外，我[a]的意思是[i]相同的
。
-
/ - Joona Palaste（pa*****@cc.helsinki.fi）-------------芬兰-------- \
\- - http://www.helsinki.fi/~palaste ---- -----------------规则！ -------- /
自行车不能自行站立，因为它是双轮胎。
- Sky Text
My teacher said that array in C is managed by preprocessor.
Preprocesser replace all array occurences (i.e. int a[10] ) with
something that I don''t understand/remember well.
What''s exactly happens with array during preprocessing/compiling stage?
Thanks in advance
/* frank */ wrote:My teacher said that array in C is managed by preprocessor.
Preprocesser replace all array occurences (i.e. int a[10] ) with
something that I don''t understand/remember well.
What''s exactly happens with array during preprocessing/compiling stage?
Thanks in advance
Well, there are two possibilities here:
1) You misunderstood what your teacher said.
2) Your teacher is horribly, horribly wrong.
I suspect you need to revisit the subject; why not *ask your teacher*?
HTH,
--ag
--
Artie Gold -- Austin, Texas
"What they accuse you of -- is what they have planned."
Artie Gold ha scritto:
1) You misunderstood what your teacher said.
If I have this array:
a[i]
the preprocessor translate it, into *(a+i)
a is translated as costant.
Thanks
/* frank */ <__*******@despammed.com> scribbled the following:Artie Gold ha scritto:1) You misunderstood what your teacher said.If I have this array: a[i] the preprocessor translate it, into *(a+i) a is translated as costant.
The preprocessor doesn''t translate it into anything. Its job is done at
that stage and it has exited stage left. [] is a full-blown operator in
C. Its behaviour is defined as the * (indirection) and + (addition)
operators combined so that a[i] means *(a+i). Also, i[a] means the same
thing as a[i].
--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"A bicycle cannot stand up by itself because it''s two-tyred."
- Sky Text
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