我用 1 个非参数构造函数、1 个参数构造函数、2 个复制构造函数、1 个赋值运算符和 1 个加号运算符编写了一个简单的 C++ 类示例。
class Complex {
protected:
float real, img;
public:
Complex () : real(0), img(0) {
cout << "Default constructor\n";
}
Complex (float a, float b) {
cout << "Param constructor" << a << " " << b << endl;
real = a;
img = b;
}
// 2 copy constructors
Complex( const Complex& other ) {
cout << "1st copy constructor " << other.real << " " << other.img << endl;
real = other.real;
img = other.img;
}
Complex( Complex& other ) {
cout << "2nd copy constructor " << other.real << " " << other.img << endl;
real = other.real;
img = other.img;
}
// assignment overloading operator
void operator= (const Complex& other) {
cout << "assignment operator " << other.real << " " << other.img << endl;
real = other.real;
img = other.img;
}
// plus overloading operator
Complex operator+ (const Complex& other) {
cout << "plus operator " << other.real << " " << other.img << endl;
float a = real + other.real;
float b = img + other.img;
return Complex(a, b);
}
float getReal () {
return real;
}
float getImg () {
return img;
}
};
我在 main 中完全像这样使用这个类:
int main() {
Complex a(1,5);
Complex b(5,7);
Complex c = a+b; // Statement 1
system("pause");
return 0;
}
结果打印为:
Param constructor 1 5
Param constructor 5 7
plus operator 5 7
Param constructor 6 12
我认为语句 1 中必须使用复制构造函数,但我真的不知道调用的是哪一个。 请告诉我是哪一个,为什么? 非常感谢
最佳答案
编译器省略了对复制构造函数的调用(实际上,两次 调用)。根据 C++11 标准的第 12.8/31 段,这是允许的(但不是强制的!),即使构造函数或析构函数有副作用:
When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the constructor selected for the copy/move operation and/or the destructor for the object have side effects. [..] This elision of copy/move operations, called copy elision, is permitted in the following circumstances (which may be combined to eliminate multiple copies):
— in a
return
statement in a function with a class return type, when the expression is the name of a non-volatile automatic object (other than a function or catch-clause parameter) with the same cv-unqualified type as the function return type, the copy/move operation can be omitted by constructing the automatic object directly into the function’s return value[...]
— when a temporary class object that has not been bound to a reference (12.2) would be copied/moved to a class object with the same cv-unqualified type, the copy/move operation can be omitted by constructing the temporary object directly into the target of the omitted copy/move
如果编译器没有省略对复制构造函数的调用,那么第一个将被选择两次,即具有以下签名的那个:
Complex( const Complex& other )
原因是:
operator +
返回的值是从临时 (Complex(a, b)
) 复制初始化的,并且只有的左值引用const
可以绑定(bind)到临时对象。如果operator +
是这样写的,事情就会不同:Complex operator+ (const Complex& other) { // ... Complex c(a, b); return c; }
在这种情况下,将调用第二个复制构造函数,因为
c
不是const
限定的并且是左值,因此它可以绑定(bind)到左值引用到非const
;main()
中的对象c
是从右值复制构造的(operator +
返回的值也是一个临时的)。无论operator +
如何返回其Complex
对象,只要它按值返回,都是如此。因此,只要不执行复制省略,就会选择第一个复制构造函数。
如果您正在使用 GCC 并想验证此行为,请尝试设置 -fno-elide-constructors
编译标志(Clang 也支持它,但 3.2 版有一个错误,我没有知道它是否已修复)。
关于c++ - 确定在 C++ 代码中调用了哪些复制构造函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16636236/