c++ - 带有 Getter 函数的 C++ 中的错误

Question

当我尝试显示一些信息时，我遇到了一个奇怪的错误。这是我的类(class)代码的一部分:

class Vertex
{
public:
    Vertex(int name)
    {
        name = name;
        pred = NULL;
        color = 'w';
        desc = 1000000;
    }

    int getName() {return name;} 
    char getColor() {return color;} 
    Vertex* getPred() {return pred;} 
    int getDesc() {return desc;}

    void setColor(char c) {color = c;} 
    void setPred(Vertex *p) {pred = p;}
    void setDesc(int d) {desc = d;}

private:
    int name;
    char color;
    Vertex* pred;
    int desc;
};

我相信这没问题......但是当我在我的主要功能中这样做时(我相信它也没问题......):

for(int z = 1; z <= nsomething; z++){
    Vertex v(z);
    cout << "from z=" << z << " is vertex name: " << v.getName() << endl;
    cout << "from z=" << z << " is vertex color: " << v.getColor() << endl;
    Vertex *vp = &v;
    cout << "from z=" << z << " is vertex name: " << vp->getName() << endl;
    cout << "from z=" << z << " is vertex color: " << vp->getColor() << endl;
}

我明白了:

from z=1 is vertex name: 1854338496
from z=1 is vertex color: w
from z=1 is vertex name: 1854338496
from z=1 is vertex color: w
from z=2 is vertex name: 1854338496
from z=2 is vertex color: w
from z=2 is vertex name: 1854338496
from z=2 is vertex color: w

有人可以帮助我，告诉我为什么只有 getName() 会发生这种情况吗？

提前致谢!

Answer 1

改变构造函数

Vertex(int name)
{
    name = name;
    pred = NULL;
    color = 'w';
    desc = 1000000;
}

以下方式

Vertex(int name)
{
    this->name = name;
    pred = NULL;
    color = 'w';
    desc = 1000000;
}

或

Vertex(int name)
{
    Vertex::name = name;
    pred = NULL;
    color = 'w';
    desc = 1000000;
}

或

Vertex(int name) : name( name )
{
    pred = NULL;
    color = 'w';
    desc = 1000000;
}

或

Vertex(int name) : name( name ), pred( NULL ), color( 'w' ), desc( 1000000 )
{
}

否则将隐藏类的数据成员name 的局部变量name 分配给它自己。

此外，最好使用限定符 const 声明所有 getter。例如

int getName() const {return name;} 
char getColor() const {return color;} 
Vertex* getPred() const {return pred;} 
int getDesc() const {return desc;}