C ++奇怪的构造函数行为 [英] C++ Strange constructor behaviour
问题描述
有人可以向我解释复杂a 和复杂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,复杂a 会调用其默认构造函数复杂b(); 不会和复杂c (0,0); 会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.
这是怎么回事?我想,复杂的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
复杂b();
是函数声明。这是没有参数并返回 Complex
对象的函数。
Complex b();
is function declaration. That is function taking no arguments and returning Complex
object.
这是非常常见的错误,并有自己的名称: most vexing parse
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
C++11 helped with this issue by introducing uniform initialization syntax
Complex b{};
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