int a [17]; int b = -1; / * a [17]与ANSI C中保证的b相同？ * / [英] int a[17]; int b = -1; /* a[17] same as b guaranteed in ANSI C? */
问题描述
在代码中
int a [17];
int b = -1;
ANSI C保证b ;位于a [16]
之后的记忆中，以便a [17]是指b？
谢谢。
bq
In the code
int a[17];
int b = -1;
does ANSI C guarantee that "b" is located in memory right after "a[16]"
so that "a[17]" refers to "b"?
Thanks.
bq
推荐答案
文章< 6c ************************** @ posting.google.com>，
bq < FO ********** @ yahoo.com>写道：
In article <6c**************************@posting.google.com >,
bq <fo**********@yahoo.com> wrote:
在代码中
int a [17];
int b = -1;
ANSI C保证b位于a [16]之后的记忆中，以便a [17]是指b？
In the code
int a[17];
int b = -1;
does ANSI C guarantee that "b" is located in memory right after "a[16]"
so that "a[17]" refers to "b"?
我没有在C标准中看到过这个
类别的任何保证。
另外，在最近的
版本的gcc上测试你的a和b的地址，结果证明b只是_before_ a在
内存地址空间。其他编译器可能表现不同。
-
Rouben Rostamian
I have not seen reference in the C standard to any guarrantee of this
sort.
Additionally, testing the addresses of your a and b on a recent
version of gcc, it turns out that b comes just _before_ a in the
memory address space. Other compilers may behave differently.
--
Rouben Rostamian
fo ********** @ yahoo.com （bq）写在
新闻：6c ************************** @ posting.google.c om：
fo**********@yahoo.com (bq) wrote in
news:6c**************************@posting.google.c om:
在代码中，
int a [17];
int b = -1;
ANSI C保证b为b。位于a [16]之后的记忆中，以便a [17]是指b？
In the code
int a[17];
int b = -1;
does ANSI C guarantee that "b" is located in memory right after "a[16]"
so that "a[17]" refers to "b"?
No.
-
- 马克 - >
-
No.
--
- Mark ->
--
2003年12月18日08:51:00 -0800， fo ********** @ yahoo.com （bq）写道：
On 18 Dec 2003 08:51:00 -0800, fo**********@yahoo.com (bq) wrote:
In代码
int a [17];
int b = -1;
ANSI C保证b位于a [16]之后的记忆中，以便a [17]是指b？
谢谢。
bq
In the code
int a[17];
int b = -1;
does ANSI C guarantee that "b" is located in memory right after "a[16]"
so that "a[17]" refers to "b"?
Thanks.
bq
号实际上，地狱，不！ ;不要那样做。
-
Al Balmer
Balmer Consulting
再**************** ********@att.net
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