指向会员的指针??? [英] Pointer to member-of-member???
问题描述
我不能为我的生活看到当
成员实际上是嵌入式结构的一部分时如何做指针成员。
也就是说，如果我有：
struct S1 {int a; };
struct S2 {S1 s; int b; };
如何在S2中获得指向a的指针？
显然我可以用b指向b >
int S2：* p =& S2 :: b;
但看似明显的语法
int S2 :: * p =& S2 :: sa;
不起作用 - 因为S2 :: sa不是/ qualified-id / - 而且
做了我能想到的其他事情。
毕竟，如果我有一个S2结构，那绝对没有区别
$ b来自编译器在s2.sa和s2.b之间的POV的$ b，那么为什么我不能
能够有一个构造s2。* p相当于前一种情况？
目前，我唯一发现的是一个相当令人讨厌的解决方法
，其中包含offsetof（）宏和大量的强制转换。也就是说，
size_t offset = offsetof（S2，sa）;
*（int *）（（char *）& s2 +偏移）= 123;
呃： - （（（
感激地收到任何更好的建议！
Steve Rencontre写道：我不能为我的生活看到如何做指针指向会员时
成员实际上是嵌入式结构的一部分。
也就是说，如果我有：
struct S1 {int a;};
struct S2 {S1 s; int b;};
如何在S2中获得指向a的指针？
显然我可以用b指向
int S2：* p =& S2 :: b;
但看似明显的语法
int S2 :: * p =& S2 :: sa;
void f（）
{
int S1 :: * p =& S1 :: a;
S2 s2;
（s2.s）。* a = 7; // parens因为我永远无法记住。* precedence
}
Steve Rencontre写道：我不能为我的生活看到如何做指针 - 当
成员实际上是嵌入式结构的一部分时成为成员。
也就是说，如果我有：
struct S1 {int a;结构S2 {S1 s; int b; }
如何在S2中获得指向a的指针？
好吧，a是一个int，所以int有什么问题？？
显然我可以用b指向b
int S2：* p =& S2 :: b;
你能吗？
但看似明显的语法
int S2 :: * p =& S2 :: s.a;
对我来说不是那么明显。有没有会员功能。
不起作用 - 因为S2 :: sa不是/ qualified-id / - 而且
不做任何我能做的事情想想。
毕竟，如果我有一个S2结构，那么与s2.sa和s2.b之间的编译器POV完全没有区别，那么为什么呢？我不应该能够拥有一个构造s2。* p相当于前一种情况吗？
目前，我唯一发现的是一个相当令人讨厌的解决方法
使用offsetof（）宏和大量的强制转换。也就是说，
size_t offset = offsetof（S2，sa）;
*（int *）（（char *）& s2 + offset）= 123;
呃： - （（
感激收到任何更好的建议！
有什么问题：
int * s2s1a =& s2.sa;
Ben Pope
-
我不只是一个数字。对于很多人来说，我被称为字符串...
Steve Rencontre写道：我不能为我的生活看到当
成员实际上是嵌入式结构的一部分时如何做指针指向成员。
就是说，如果我有：
struct S1 {int a;};
struct S2 {S1 s; int b;};
如何获得指向a的指针在一个S2？
显然我可以用b指向b：S2：* p =& S2 :: b;
这个是不是有效的C ++语法。
但看似明显的语法
int S2 :: * p =& S2 :: sa;
？
类/结构的数据成员只能通过一个实例
该对象：
S2 s2;
int * p =& s2.sa ;
您是否对指向会员职能的指针感到困惑？
-
Ian Collins。
I can''t for the life of me see how to do pointer-to-member when the
member is actually part of an embedded structure.
That is, if I have:
struct S1 { int a; };
struct S2 { S1 s; int b; };
how can I get a pointer to the a in an S2?
Obviously I can point at the b with
int S2:*p = &S2::b;
But the seemingly obvious syntax
int S2::*p = &S2::s.a;
doesn''t work - because S2::s.a is not a /qualified-id/ - and neither
does anything else I can think of.
After all, if I have an S2 structure, there''s absolutely no difference
from the compiler''s POV between s2.s.a and s2.b, so why should I not be
able to have a construct s2.*p equivalent to the former case?
At the moment, the only thing I have found is a rather yucky workaround
with offsetof() macros and lots of casts. That is,
size_t offset = offsetof (S2, s.a);
*(int *) ((char *) &s2 + offset) = 123;
Ugh :-(((
Any better suggestions gratefully received!
Steve Rencontre wrote:I can''t for the life of me see how to do pointer-to-member when the
member is actually part of an embedded structure.
That is, if I have:
struct S1 { int a; };
struct S2 { S1 s; int b; };
how can I get a pointer to the a in an S2?
Obviously I can point at the b with
int S2:*p = &S2::b;
But the seemingly obvious syntax
int S2::*p = &S2::s.a;
void f()
{
int S1::*p = &S1::a;
S2 s2;
(s2.s).*a = 7; // parens because I can never remember .* precedence
}
Steve Rencontre wrote:I can''t for the life of me see how to do pointer-to-member when the
member is actually part of an embedded structure.
That is, if I have:
struct S1 { int a; };
struct S2 { S1 s; int b; };
how can I get a pointer to the a in an S2?
Well, a is an int, so what''s wrong with an int*?
Obviously I can point at the b with
int S2:*p = &S2::b;
Can you?
But the seemingly obvious syntax
int S2::*p = &S2::s.a;
Not so obvious to me. There''s aren''t member functions.
doesn''t work - because S2::s.a is not a /qualified-id/ - and neither
does anything else I can think of.
After all, if I have an S2 structure, there''s absolutely no difference
from the compiler''s POV between s2.s.a and s2.b, so why should I not be
able to have a construct s2.*p equivalent to the former case?
At the moment, the only thing I have found is a rather yucky workaround
with offsetof() macros and lots of casts. That is,
size_t offset = offsetof (S2, s.a);
*(int *) ((char *) &s2 + offset) = 123;
Ugh :-(((
Any better suggestions gratefully received!
What''s wrong with:
int* s2s1a = &s2.s.a;
Ben Pope
--
I''m not just a number. To many, I''m known as a string...
Steve Rencontre wrote:I can''t for the life of me see how to do pointer-to-member when the
member is actually part of an embedded structure.
That is, if I have:
struct S1 { int a; };
struct S2 { S1 s; int b; };
how can I get a pointer to the a in an S2?
Obviously I can point at the b with
int S2:*p = &S2::b;
This isn''t valid C++ syntax.
But the seemingly obvious syntax
int S2::*p = &S2::s.a;
?
Data members of a class/struct an only be accesses through an instance
of that object:
S2 s2;
int* p = &s2.s.a;
Are you confused with pointers to member functions?
--
Ian Collins.
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