Question

class a
{
  private:
    b *b_obj;
  public:
    void set(int);
};

a::a()
{
  b_obj = new b;
}

a::set(int s)
{
  b_obj->c = s;
 }

class b
{
  public:
    int c;
};

is this code valid? if no, how do i make b_obj of a particular object (say a_obj) of class a ,modifiable in another class c...if a_obj i created in another class d....i am scared of a_obj going out of scope in class c.

hope you understand my question. thanks a lot for taking the time to read my post

Answer 1

该守则有效。 www.un.org/spanish/ga/president 这里是修订后的法典,还有一项试验:

class b
{
  public:
    int c;
};

class a
{
  private:
    b *b_obj;
  public:
    a();
    void set(int);
};

a::a()
{
  b_obj = new b;
}

void a::set(int s)
{
  b_obj->c = s;
}

int main()
{
    a my_a;
    my_a.set(42);
}

现在就是因为该守则是valid<>。页: 1

You don t initialize c when default-constructing b.

You use raw pointers to dyanamically-allocated b s. Use automatic variables instead, whenever possible. Among the the reasons for this are ...

You never delete the b you new ed in a a constructor. This results in a memory leak. If you had avoided the use of dynamic allocation in the first place, this would not be an issue.

Answer 2

水井

*b_obj = *other_b_obj;// equality of objects values
b_obj = other_b_obj; // equality for pointer values

假定b_obj为物体和other_b_obj为点:

b_obj = *other_b_obj;

假设相反:

b_obj = &other_b_obj;

两者都是要点:

b_obj = other_b_obj;

对于您的最后一个问题,new对成点人并非强制性的。点人可指出物。然而,如果你想要指明新物体,则使用<条码>新关键词,试图制造新物体,并将新物体地址退回。

Answer 3

is this code valid?

该代码为几乎所有有效,但你不敢说明既定职能的返回类型:

void a::set(int s)

但是,既定职能将真正改变<代码>c > 成员<编码>b_obj。它是公开的,因此汇编者将允许这样做。

http://ideone.com/4uDnE>rel=“nofollow> BTW,为什么没有你试图放弃吗?

除此以外,在你重新研究时,看构造者、骑士、派任经营人,并妥善执行,以释放标的<代码>b_obj。甚至可以为此使用<条码>。

另外,我也建议你使用一个班级的“低地编码”()。

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