表达式必须具有类类型
一段时间以来我都没有用C ++编写代码,当我尝试编译这个简单的代码片段时被卡住了:
I have't coded in c++ for some time and I got stuck when I tried to compile this simple snippet:
class A
{
public:
void f() {}
};
int main()
{
{
A a;
a.f(); // works fine
}
{
A *a = new A();
a.f(); // this doesn't
}
}
它是一个指针,因此请尝试:
It's a pointer, so instead try:
a->f();
基本上是运算符。
(用于访问对象和引用使用对象的字段和方法),因此:
Basically the operator .
(used to access an object's fields and methods) is used on objects and references, so:
A a;
a.f();
A& ref = a;
ref.f();
如果您使用的是指针类型,则必须先取消引用以获得引用:
If you have a pointer type, you have to dereference it first to obtain a reference:
A* ptr = new A();
(*ptr).f();
ptr->f();
a-&b; $
表示法是通常只是(* a).b
的简写。
The a->b
notation is usually just a shorthand for (*a).b
.
operator->
可以重载,这在智能指针中尤为明显。当您正在使用智能指针时>,那么您还可以使用->
来引用指向的对象:
The operator->
can be overloaded, which is notably used by smart pointers. When you're using smart pointers, then you also use ->
to refer to the pointed object:
auto ptr = make_unique<A>();
ptr->f();