某人可以向我解释<代码>(*this)点,并表明如何将其用作同一类别另一物体的物体。 我不理解的是,你将如何在两个不同目标的成员职能中以相同的变数提及两个数字。 例如,增加两个数字。
a.Multiply(b);
//....
Numbers::Numbers Multiply(Numbers auggend)
{
}
某人可以向我解释<代码>(*this)点,并表明如何将其用作同一类别另一物体的物体。 我不理解的是,你将如何在两个不同目标的成员职能中以相同的变数提及两个数字。 例如,增加两个数字。
a.Multiply(b);
//....
Numbers::Numbers Multiply(Numbers auggend)
{
}
this pointer:
每一物体的成员职能可进入一个标明物体本身的点名this
。 当我们称之为成员职能时,它就存在,其价值为>、
。 使用<条码>该条码>指任何成员均可找到其所属物体的地址。 也可使用该数据获取其标的中的数据。 例:
void setdata(int ii)
{
i=ii; // one way to set data
this->i=ii; // another way to set data
}
http://www.un.org/Depts/DGACM/index_french.htm
(* )是你宣布某类人的一个要点。 举例来说,如果数字类别具有数据价值“价值”:
a.Multiply(b);
....
Numbers::Numbers Multiply(Numbers auggend)
{
return (this->value) * (auggend.value);
}
首先,你的例子并不正确。 我认为应该:
Numbers Numbers::Multiply(Numbers auggend)
{
}
如上所述,<代码>该代码>只是一个类型的变量:<代码>Numbers * const,标明了你的成员变量。 因此,根据您的榜样,a.Multiply(b)
,this
将具有&a
的价值。
请允许我说,你还有其他一些职能,采用了<代码>。 编号*:
void DoSomething(Numbers *num);
之后,你可以把这一职能称作this
:
Numbers Numbers::Multiple(Numbers auggend)
{
DoSomething(this);
DoSomething(&auggend);
}
class A{
int num;
void foo(int num)
{
num = 10; //local variable num is set to 10
this->num = 10 ; //class member num is set to 10
}
};
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