C++ 奇怪的构造函数行为
谁能给我解释一下Complex a;和Complex b();之间的区别吗?
Can anybody explain to me the difference between Complex a; and Complex b();?
#include<iostream>
class Complex
{
public:
Complex()
{
std::cout << "Complex Constructor 1" << std::endl;
}
Complex(float re, float im)
{
std::cout << "Complex Constructor 2" << std::endl;
}
~Complex()
{
std::cout << "Complex Destructor" << std::endl;
}
};
int main()
{
Complex a;
std::cout << "--------------------------" << std::endl;
Complex b();
std::cout << "--------------------------" << std::endl;
Complex c(0,0);
std::cout << "--------------------------" << std::endl;
return 0;
}
输出:
Complex Constructor 1
--------------------------
--------------------------
Complex Constructor 2
--------------------------
Complex Destructor
Complex Destructor
如您所见,Complex a; 确实调用了它的默认构造函数,Complex b(); 没有,Complex c(0,0); 调用重载的构造函数.
As you can see, Complex a; does call its default constructor, Complex b(); doesn't and Complex c(0,0); calls an overloaded constructor.
这里发生了什么?我想,Complex b(); 会创建一个堆栈变量并调用它的默认构造函数来初始化它?
What is going on here? I thought, that Complex b(); would create a stack-variable and call it's default constructor to initialize it?
推荐答案
Complex b();
是函数声明.这是不带参数并返回 Complex
对象的函数.
Complex b();
is function declaration. That is function taking no arguments and returning Complex
object.
这是一个很常见的错误并且有自己的名字:最令人头疼的解析
This is very common mistake and has its own name: most vexing parse
C++11 通过引入统一初始化语法帮助解决了这个问题
C++11 helped with this issue by introducing uniform initialization syntax
Complex b{};
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