为什么在此代码中不调用复制构造函数 [英] Why is copy constructor not being called in this code
问题描述
那么为什么不在 const Integer operator +(const Integer& rv)函数中调用Copy构造函数。是因为RVO。 如果是,我需要做什么来防止它?
#include< iostream&
using namespace std;
class Integer {
int i;
public:
Integer(int ii = 0):i(ii){
cout< Integer()<< endl;
}
Integer(const Integer& I){
cout< Integer(const Integer&)<< endl;
}
〜Integer(){
cout< 〜Integer()<< endl;
}
const整数运算符+(const Integer& rv)const {
cout< operator +<< endl;
Integer I(i + rv.i);
I.print();
return I;
}
Integer& operator + =(const Integer& rv){
cout< operator + =<< endl;
i + rv.i;
return * this;
}
void print(){
cout< i:< i<< endl;
}
};
int main(){
cout<< 内置tpes:< endl;
int i = 1,j = 2,k = 3;
k + = i + j;
cout<< user-defined types:<< endl;
Integer ii(1),jj(2),kk(3);
kk + = ii + jj;
}
我会收到错误构造函数。我期待复制构造函数,当operator +返回时被调用。以下是程序的输出
内置tpes:
用户定义的类型:
Integer ()
Integer()
Integer()
operator +
Integer()
i:3 //预期复制构造函数在此之后调用
operator +
〜Integer()
〜Integer()
〜Integer()
〜Integer()
解决方案
是因为RVO。如果是,我需要做什么来防止它?
是的。但它没有被调用,因为编译器返回值优化。
如果您使用GCC,请使用
-fno-elide-constructors
选项来避免它。
-fno-elide-constructors / p>
C ++标准允许实现省略创建只用于初始化同一类型的另一个对象的临时。指定此选项将禁用该优化,并强制G ++在所有情况下调用复制构造函数。
So why is Copy constructor not being invoked in "const Integer operator+(const Integer &rv)" function. Is it because of RVO. If Yes what do I need to do to prevent it?
#include <iostream> using namespace std; class Integer { int i; public: Integer(int ii = 0) : i(ii) { cout << "Integer()" << endl; } Integer(const Integer &I) { cout << "Integer(const Integer &)" << endl; } ~Integer() { cout << "~Integer()" << endl; } const Integer operator+(const Integer &rv) const { cout << "operator+" << endl; Integer I(i + rv.i); I.print(); return I; } Integer &operator+=(const Integer &rv) { cout << "operator+=" << endl; i + rv.i; return *this; } void print() { cout << "i: " << i << endl; } }; int main() { cout << "built-in tpes:" << endl; int i = 1, j = 2, k = 3; k += i + j; cout << "user-defined types:" << endl; Integer ii(1), jj(2), kk(3); kk += ii + jj; }
I do get an error If I'll comment out copy constructor. I'm expecting copy constructor to be called when operator+ returns. Following is the output of the program
built-in tpes: user-defined types: Integer() Integer() Integer() operator+ Integer() i: 3 // EXPECTING Copy Constructor to be called after this operator+= ~Integer() ~Integer() ~Integer() ~Integer()
解决方案Is it because of RVO. If Yes what do I need to do to prevent it?
Yes. But it didn't get called because of Return Value Optimization by the compiler.
If you're using GCC, then use
-fno-elide-constructors
option to avoid it.GCC 4.6.1 manual says,
-fno-elide-constructors
The C++ standard allows an implementation to omit creating a temporary which is only used to initialize another object of the same type. Specifying this option disables that optimization, and forces G++ to call the copy constructor in all cases.
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