数组列表？ [英] Array List?
问题描述
来自C和Java的* nix我用CList和MSVC ++搞砸了我的元素，但我认为我的问题主要是语法上的。
我有一个我称之为''varray'的ADT，可以在给定索引的数组中返回指向
arbirary大小元素的指针，它将分配
内存在必要时支持它：
struct varray * tests = varray_new（sizeof（struct test））;
struct test * t =（struct test *）varray_get（tests，50））;
重点是我不必为* t分配内存。支持数组中的每个元素都是sizeof（struct test）。
我应该使用C ++中的等效ADT吗？
谢谢，
迈克
ref： http://www.ioplex.com/~miallen/libmba/dl/src/varray.c
" Alf P. Steinbach" <人*** @ start.no>在消息中写道
news：3f **************** @ News.CIS.DFN.DE ...< blockquote class =post_quotes>我有一个我称之为''varray'的ADT，可以在给定索引的数组中返回指向
anarbirary大小元素的指针，它将分配
thememory to如有必要，请回复：
struct varray * tests = varray_new（sizeof（struct test））;
struct test * t =（struct test *）varray_get（tests，50））;
关键是我没有为* t分配内存。支持数组中的每个元素
都是sizeof（struct test）。
我应该使用C ++中的等效ADT吗？
签出标准容器，std :: vector，std :: list等。
你的意思是这样的？这是合法的吗？
#include< iostream>
#include< list>
using namespace std;
struct foo {
int i;
short s;
};
main（）
{
list< struct foo> L;
struct foo f;
fi = 10;
fs = 5;
L.push_back（f）;
fi = 30;
fs = 15;
L.push_front（f）;
fi = 100;
fs = 50;
L.insert（++ L.begin（），f）;
fi = 101;
fs = 51;
L.push_back（f）;
fi = 102;
fs = 52;
L.push_back（f）;
list< struct foo> :: iterator i;
for（i = L.begin（）; i！= L.end（）; ++ i）cout<< " I =" << i-> i<< "，S =" << i-> s
<< " " ;;
cout<<结束;
返回0;
}
c：\ miallen \p> list
i = 30，s = 15 i = 100，s = 50 i = 10，s = 5 i = 101，s = 51 i = 102，s = 52
在2003年9月8日星期一14:10:16 -0700，Michael B. Allen < MB ***** @ ioplex.com>写道：
" Alf P. Steinbach" <人*** @ start.no>在消息中写道
新闻：3f **************** @ News.CIS.DFN.DE ...>我有一个我称之为''varray'的ADT，可以在给定索引的数组中返回一个指向> arbirary大小元素的指针，如果需要，它将分配>内存来支持它：
>
> struct varray * tests = varray_new（sizeof（struct test））;
> struct test * t =（struct test *）varray_get（tests，50））;
>
>重点是我没有为* t分配内存。每个元素>支持数组是sizeof（结构测试）。
>
>我应该使用C ++中的等效ADT吗？
检查标准容器，std： ：vector，std :: list等。
你的意思是这样的吗？
好。我只会评论改进的机会。
这是合法的吗？
是的。
#include< iostream>
#include< list>
使用namespace std;
防止你出于自己的目的使用''list'这样的名字。
struct foo {
int i;
短片;
};
构造函数可以大大减轻事情，避免例如假人
变量如下面的''f''，以及作业。
main（）
''main''必须有返回类型''int''。
见[ http://www.research.att.com/~bs/bs_faq2.html#void-main] 。
{
list< struct foo>升;
无需重复单词''struct''（与C相反，在C ++中a />
''struct''类似于任何类型其他）。
避免使用所有大写名称可以帮助避免与
宏发生名称冲突。
struct foo f ;
fi = 10;
fs = 5;
L.push_back（f）;
fi = 30;
fs = 15;
L.push_front（f）;
fi = 100;
fs = 50;
L.insert（++ L.begin（），f）;
fi = 101;
fs = 51;
L.push_back（f）;
fi = 102;
fs = 52;
L.push_back（f）;
list< struct foo> :: iterator i;
for（i = L.begin（）; i！= L.end（）; ++ i）cout< < " I =" << i-> i<< "，S =" << i-> s
<< " " ;;
cout<<结束;
返回0;
}
考虑
#include< iostream>
#include< list>
struct Foo
{
int i;
短片;
Foo（int iValue，短sValue）：i（iValue），s（sValue）{}
};
int main（）
{
std :: list< Foo>列表;
list.push_back（Foo（10,5））;
list.push_front（Foo（30,15））;
list.insert（++ list.begin（），Foo（100,50））;
list.push_back（Foo（101,51））;
list.push_back（Foo（102,52））;
std :: list< Foo> :: iterator i;
for（i = list.begin（）; i！= list.end（）; ++ i）
{
std :: cout<< " I =" << i-> i<< "，S =" << i-> s<< " " << std :: endl;
}
}
c：\ miallen \p> list
i = 30，s = 15 i = 100，s = 50 i = 10，s = 5 i = 101，s = 51 i = 102，s = 52
< br>
" Alf P. Steinbach" <人*** @ start.no>在留言中写道
news：3f **************** @ News.CIS.DFN.DE ...＃包括< iostream>
#include< list>
struct Foo
{
int i;
短片;
Foo（int iValue，short sValue）：i（iValue），s（sValue）{}
};
int main（）
{
std ::列表<富> list;
list.push_back（Foo（10,5））;
"
有趣。那是构造函数吗？这个和
之间的关系是一个类的构造函数吗？这会退化成一个类还是这个
只是一些特定于结构的额外符号？
在实践中，人们最终是否完全使用类结构？或者是
使用像这样的结构有什么优势？
让我对C ++感到困扰的一件事就是你不能确定内存中的内容
（例如你不能序列化，检查或部分复制占用的内存<由于害怕破坏它而被一个班级
）。明确的记忆操作是什么
使C很棒（是的，也很危险）。
谢谢，
Mike
Coming from C and Java on *nix I''m a little out of my element messing around
with CList and MSVC++ but I think my issues are largely syntactic.
I have an ADT that I use called a ''varray'' that can return a pointer to an
arbirary sized element in an array given an index and it will allocate the
memory to back it if necessary:
struct varray *tests = varray_new(sizeof(struct test));
struct test *t = (struct test *)varray_get(tests, 50));
The point being I don''t have to allocate the memory for *t. Each element in
the backing array is sizeof(struct test).
Is there an equivalent ADT in C++ that I should use?
Thanks,
Mike
ref: http://www.ioplex.com/~miallen/libmba/dl/src/varray.c
"Alf P. Steinbach" <al***@start.no> wrote in message
news:3f****************@News.CIS.DFN.DE...I have an ADT that I use called a ''varray'' that can return a pointer to anarbirary sized element in an array given an index and it will allocate thememory to back it if necessary:
struct varray *tests = varray_new(sizeof(struct test));
struct test *t = (struct test *)varray_get(tests, 50));
The point being I don''t have to allocate the memory for *t. Each element inthe backing array is sizeof(struct test).
Is there an equivalent ADT in C++ that I should use?
Check out the standard containers, std::vector, std::list and so on.
You mean like this? Is this legit?
#include <iostream>
#include <list>
using namespace std;
struct foo {
int i;
short s;
};
main()
{
list<struct foo> L;
struct foo f;
f.i = 10;
f.s = 5;
L.push_back(f);
f.i = 30;
f.s = 15;
L.push_front(f);
f.i = 100;
f.s = 50;
L.insert(++L.begin(),f);
f.i = 101;
f.s = 51;
L.push_back(f);
f.i = 102;
f.s = 52;
L.push_back(f);
list<struct foo>::iterator i;
for(i=L.begin(); i != L.end(); ++i) cout << "i=" << i->i << ",s=" << i->s
<< " ";
cout << endl;
return 0;
}
c:\miallen\p>list
i=30,s=15 i=100,s=50 i=10,s=5 i=101,s=51 i=102,s=52
On Mon, 8 Sep 2003 14:10:16 -0700, "Michael B. Allen" <mb*****@ioplex.com> wrote:
"Alf P. Steinbach" <al***@start.no> wrote in message
news:3f****************@News.CIS.DFN.DE...>I have an ADT that I use called a ''varray'' that can return a pointer toan >arbirary sized element in an array given an index and it will allocatethe >memory to back it if necessary:
>
> struct varray *tests = varray_new(sizeof(struct test));
> struct test *t = (struct test *)varray_get(tests, 50));
>
>The point being I don''t have to allocate the memory for *t. Each elementin >the backing array is sizeof(struct test).
>
>Is there an equivalent ADT in C++ that I should use?
Check out the standard containers, std::vector, std::list and so on.
You mean like this?
Good. I''ll comment only on opportunities for improvement.
Is this legit?
Yep.
#include <iostream>
#include <list>
using namespace std;
Prevents you from using names like ''list'' for your own purposes.
struct foo {
int i;
short s;
};
A constructor can ease things considerably, avoiding e.g. dummy
variables like ''f'' below, as well as the assignments.
main()
''main'' must have return type ''int''.
See [http://www.research.att.com/~bs/bs_faq2.html#void-main].
{
list<struct foo> L;
No need to repeat the word ''struct'' (in contrast to C, in C++ a
''struct'' is a type like any others).
Avoiding all uppercase names can help avoid name conflicts with
macros.
struct foo f;
f.i = 10;
f.s = 5;
L.push_back(f);
f.i = 30;
f.s = 15;
L.push_front(f);
f.i = 100;
f.s = 50;
L.insert(++L.begin(),f);
f.i = 101;
f.s = 51;
L.push_back(f);
f.i = 102;
f.s = 52;
L.push_back(f);
list<struct foo>::iterator i;
for(i=L.begin(); i != L.end(); ++i) cout << "i=" << i->i << ",s=" << i->s
<< " ";
cout << endl;
return 0;
}
Consider
#include <iostream>
#include <list>
struct Foo
{
int i;
short s;
Foo( int iValue, short sValue ): i( iValue ), s( sValue ) {}
};
int main()
{
std::list<Foo> list;
list.push_back( Foo( 10, 5 ) );
list.push_front( Foo( 30, 15 ) );
list.insert( ++list.begin(), Foo( 100, 50 ) );
list.push_back( Foo( 101, 51 ) );
list.push_back( Foo( 102, 52 ) );
std::list<Foo>::iterator i;
for( i = list.begin(); i != list.end(); ++i )
{
std::cout << "i=" << i->i << ",s=" << i->s << " " << std::endl;
}
}
c:\miallen\p>list
i=30,s=15 i=100,s=50 i=10,s=5 i=101,s=51 i=102,s=52
"Alf P. Steinbach" <al***@start.no> wrote in message
news:3f****************@News.CIS.DFN.DE...#include <iostream>
#include <list>
struct Foo
{
int i;
short s;
Foo( int iValue, short sValue ): i( iValue ), s( sValue ) {}
};
int main()
{
std::list<Foo> list;
list.push_back( Foo( 10, 5 ) );
"
Interesting. Is that a constructor? Is there a relationship between this and
the constructor of a class? Does this degenerate into a class or is this
just some extra notation specific to structures?
In practice do folks just end up using classes over structs entirely? Or is
there an advantage to using a struct like this? One of the things that
bothered me about C++ that you cannot not be certain what was in memory
(e.g. you cannot serialize, inspect, or partially copy the memory occupied
by a class for fear of breaking it). Explicit memory manipluation is what
makes C great (yes, and dangerous).
Thanks,
Mike
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