class Node
{
public:
Node *parent; // used during the search to record the parent of successor nodes
Node *child; // used after the search for the application to view the search in reverse
float g; // cost of this node + it's predecessors
float h; // heuristic estimate of distance to goal
float f; // sum of cumulative cost of predecessors and self and heuristic
Node() :
parent( 0 ),child( 0 ),g( 0.0f ),h( 0.0f ),f( 0.0f )
{
}
UserState m_UserState;
};
为什么我们应该使用构造函数
Node() :
parent( 0 ),f( 0.0f )
{
}
代替
Node()
{
parent = null;
child = null;
g = 0.0f;
h = 0.0f;
f = 0.0f;
}
谢谢 ：)
解决方法
对于简单的旧数据(POD),这没有什么好处,但一旦开始使用引用或组合类,它会有所不同：
class Foo {
Bar bar;
public:
// construct bar from x
Foo(int x) : bar(x) { }
};
与
Foo::Foo(int x)
{
// bar is default-constructed; how do we "re-construct" it from x?
bar = x; // requires operator=(int) on bar; even if that's available,// time is wasted default-constructing bar
}
有时,一旦构造了对象,你甚至不会有一种“重新构建”对象的方式,因为类可能不支持setter或operator =. const成员肯定不能“重新构建”或重置：
class FooWithConstBar {
const Bar bar;
public:
Foo(int x) {
// bar is cast in stone for the lifetime of this Foo object
}
};
编辑：感谢@Vitus指出引用的问题.
