指针相等 [英] pointer equality
问题描述
在K& R" C ++编程语言（第2版ANSI C版）中，参考
手册指出（第7.9和7.10段）指针比较是
对于没有指向同一个对象的指针，未定义。
所以如果我们有
const char * foo =" foo" ，* bar =" bar" ;
int foobar =（foo == bar）;
是否意味着foobar未定义？
我知道关系运算符（<，< =，> =，>）不能用于指针
到不同的对象，但是等于运算符==，！=？
问候，
Ike
邮寄到ike at ibe dot nl
In K&R "The C++ programming language (2nd ANSI C edition), the reference
manual states (paragraphs 7.9 and 7.10) that pointer comparison is
undefined for pointers that do not point to the same object.
So if we have
const char * foo = "foo" , * bar = "bar" ;
int foobar = ( foo == bar ) ;
would it mean that foobar is undefined?
I knew that relational operators (<,<=,>=,>) cannot be used on pointers
to different objects, but how about equality operators ==,!= ?
Regards,
Ike
mail to ike at iae dot nl
推荐答案
2004年4月6日星期二21:25:13 GMT，Ike Naar< no **** @ nospam.invalid>写道：
On Tue, 06 Apr 2004 21:25:13 GMT, Ike Naar <no****@nospam.invalid> wrote:
在K& R" C ++编程语言（第2版ANSI C版）中，参考指南比较为手册指出（第7.9和7.10段）对于没有指向同一个对象的指针，未定义。
如果我们有
const char * foo =" foo" ，* bar =" bar" ;
int foobar =（foo == bar）;
这是否意味着foobar未定义？
你怎么看？我正在看：
指向相同类型的物体的指针（忽略任何限定符）
可以比较
我知道关系运算符（<，< =，> =，>）不能用于指针
到不同的对象
为什么不呢？
int a [10];
int * p1 =& a [3];
int * p2 =& a [4];
if（p2> p1）
...
完全合法。
，但怎么样平等运算符==，！=？
当然，要测试指针是否指向同一个对象......
-leor
问候，
Ike
给ike发送邮件给点nl
In K&R "The C++ programming language (2nd ANSI C edition), the reference
manual states (paragraphs 7.9 and 7.10) that pointer comparison is
undefined for pointers that do not point to the same object.
So if we have
const char * foo = "foo" , * bar = "bar" ;
int foobar = ( foo == bar ) ;
would it mean that foobar is undefined?
How are you reading that? I''m seeing:
"Pointers to objects of the same type (ignoring any qualifiers)
may be compared"
I knew that relational operators (<,<=,>=,>) cannot be used on pointers
to different objects
Why not?
int a[10];
int *p1 = &a[3];
int *p2 = &a[4];
if (p2 > p1)
...
Perfectly legal.
, but how about equality operators ==,!= ?
Of course, to test if the pointers point to the same object or not...
-leor
Regards,
Ike
mail to ike at iae dot nl
-
Leor Zolman - - BD软件--- www.bdsoft.com
开 - C / C ++，Java，Perl和Unix的现场培训
C ++用户：下载BD Software的免费STL错误消息解密器：
www.bdsoft.com/tools/stlfilt.html
2004年4月6日星期二21:25:13 GMT，Ike Naar< no **** @ nospam.invalid>写道：
On Tue, 06 Apr 2004 21:25:13 GMT, Ike Naar <no****@nospam.invalid> wrote:
在K& R" C ++编程语言（第2版ANSI C版）中，参考指南比较为手册指出（第7.9和7.10段）对于没有指向同一个对象的指针，未定义。
如果我们有
const char * foo =" foo" ，* bar =" bar" ;
int foobar =（foo == bar）;
这是否意味着foobar未定义？
你怎么看？我正在看：
指向相同类型的物体的指针（忽略任何限定符）
可以比较
我知道关系运算符（<，< =，> =，>）不能用于指针
到不同的对象
为什么不呢？
int a [10];
int * p1 =& a [3];
int * p2 =& a [4];
if（p2> p1）
...
完全合法。
，但怎么样平等运算符==，！=？
当然，要测试指针是否指向同一个对象......
-leor
问候，
Ike
给ike发送邮件给点nl
In K&R "The C++ programming language (2nd ANSI C edition), the reference
manual states (paragraphs 7.9 and 7.10) that pointer comparison is
undefined for pointers that do not point to the same object.
So if we have
const char * foo = "foo" , * bar = "bar" ;
int foobar = ( foo == bar ) ;
would it mean that foobar is undefined?
How are you reading that? I''m seeing:
"Pointers to objects of the same type (ignoring any qualifiers)
may be compared"
I knew that relational operators (<,<=,>=,>) cannot be used on pointers
to different objects
Why not?
int a[10];
int *p1 = &a[3];
int *p2 = &a[4];
if (p2 > p1)
...
Perfectly legal.
, but how about equality operators ==,!= ?
Of course, to test if the pointers point to the same object or not...
-leor
Regards,
Ike
mail to ike at iae dot nl
-
Leor Zolman - - BD软件--- www.bdsoft.com
开 - C / C ++，Java，Perl和Unix的现场培训
C ++用户：下载BD Software的免费STL错误消息解密器：
www.bdsoft.com/tools/stlfilt.html
在文章< Ze **************** @ typhoon.bart.nl>，
Ike Naar< no **** @ nospam.invalid>写道：
In article <Ze****************@typhoon.bart.nl>,
Ike Naar <no****@nospam.invalid> wrote:
在K& R" C ++编程语言（第2版ANSI C版）中，参考指南比较为手册指出（第7.9和7.10段）对于没有指向同一个对象的指针，未定义。
如果我们有
const char * foo =" foo" ，* bar =" bar" ;
int foobar =（foo == bar）;
这是否意味着foobar未定义？
我知道关系运算符（<，< =，> =，>）不能用于指向不同对象的指针，但是等于运算符怎么样==，！=？
In K&R "The C++ programming language (2nd ANSI C edition), the reference
manual states (paragraphs 7.9 and 7.10) that pointer comparison is
undefined for pointers that do not point to the same object.
So if we have
const char * foo = "foo" , * bar = "bar" ;
int foobar = ( foo == bar ) ;
would it mean that foobar is undefined? I knew that relational operators (<,<=,>=,>) cannot be used on pointers
to different objects, but how about equality operators ==,!= ?
< < => > =不能合法地用于指向不同对象的指针; ==
和！=可以。所以你提到的那本书是错的，或者当你读到它时你错过了一些东西。你可能想重新阅读它，看看它是否是b $ b提及关系运营商。和平等操作者。如果你是
确定你不会误解任何东西，你应该从这本书中更多地发布一些
。
< <= > >= cannot be legally used on pointers to different objects; ==
and != can. So either the book you mentioned is wrong, or you missed
something when you read it. You might want to re-read it and see if it
mentions "relational operators" and "equality operators". If you are
sure you don''t misunderstand anything, you should probably post a bit
more literally from that book.
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