关于指针的问题 [英] Question on Pointers
问题描述
各位大家好！
我有疑问，需要某种确认！
是不是
pChar [2] =''a''与
*（pChar + 2）=''a''相同！
（其中char * pChar和pChar = malloc（20））
我对第一个例子感到困惑。为什么？
零
零< ch **** ****@web.de>写道：各位大家好！
我有一个问题，需要某种确认！
是不是
pChar [2] =''a''与
相同*（pChar + 2）=''a''！
（其中char * pChar和pChar = malloc （20））
我对第一个例子感到困惑。为什么会这样？
指针和数组的等效通常会让那些新来的人感到困惑。
comp.lang.c faq有专门讨论该主题的章节：
http：/ /c-faq.com/aryptr/index.html
阅读此内容可能会回答您的问题并消除您的困惑。
-
：wq
^ X ^ Cy ^ K ^ X ^ C ^ C ^ C ^ C
"零" < CH ******** @ web.de>在消息中写道
新闻：11 ********************** @ g43g2000cwa.googlegr oups.com ...
：各位大家好！
：
：我有一个问题，需要某种确认！
：
：是不是
：
：pChar [2] =''a''与
相同： *（pChar + 2）=''a''！
是的。
即使以下是相同的：
2 [pChar] =''a''
:(其中char * pChar和pChar = malloc（20））
：
：我对第一个例子感到困惑。为什么会这样？
为什么你会感到困惑？
从指针偏移处访问元素
被视为经常需要得到自己的友好符号
a
就像a-> x，相当于（* a）.x
Ivan
-
http://ivan.vecerina.com/contact/?subject=NG_POST < - 电子邮件联系表格
Brainbench MVP for C ++<> http://www.brainbench.com
我感到困惑，因为我将pChar [2]解释为偏移地址，所以
我改变地址而不是地址后面的值。
Hello everybody!
I have a question and need some kind of confirmation!
Is it right that
pChar[2] = ''a'' is the same as
*(pChar + 2) = ''a''!
(where char * pChar and pChar = malloc(20))
I get confused with the first example. Why is that?
Zero
Zero <ch********@web.de> wrote:Hello everybody!
I have a question and need some kind of confirmation!
Is it right that
pChar[2] = ''a'' is the same as
*(pChar + 2) = ''a''!
(where char * pChar and pChar = malloc(20))
I get confused with the first example. Why is that?
Pointer and array equivalance often can be confusing to those new to C.
The comp.lang.c faq has a chapter dedicated to this subject :
http://c-faq.com/aryptr/index.html
Reading this might answer your question and clear up your confusion.
--
:wq
^X^Cy^K^X^C^C^C^C
"Zero" <ch********@web.de> wrote in message
news:11**********************@g43g2000cwa.googlegr oups.com...
: Hello everybody!
:
: I have a question and need some kind of confirmation!
:
: Is it right that
:
: pChar[2] = ''a'' is the same as
: *(pChar + 2) = ''a''!
yes.
Even the following is equivalent:
2[pChar] = ''a''
: (where char * pChar and pChar = malloc(20))
:
: I get confused with the first example. Why is that?
Why are you getting confused?
Accessing an element at an offset from a pointer
was seen as a frequent enough need to deserve
a friendly notation of its own.
Just as a->x, which is equivalent to (*a).x
Ivan
--
http://ivan.vecerina.com/contact/?subject=NG_POST <- email contact form
Brainbench MVP for C++ <> http://www.brainbench.com
I get confused because I interpret pChar[2] as an address to offset, so
I change the address not the value behind the address.
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