传递对引用的引用 [英] Pass a reference to a reference
本文介绍了传递对引用的引用的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我认为在 C++ 中传递对引用的引用是非法的.但是,当我运行此代码时,它没有给我任何错误.
I think it's illegal to pass a reference to a reference in C++.However ,when I run this code it gives me no error.
void g(int& y)
{
std::cout << y;
y++;
}
void f(int& x)
{
g(x);
}
int main()
{
int a = 34;
f(a);
return 0;
}
是不是形式参数G()有资格作为对参考的参考 ??
Doesn't the formal parameter of g() qualify as a reference to a reference ??
推荐答案
1) 将引用传递给引用并没有错(它是移动构造函数和移动赋值运算符使用的 - 尽管它实际上被称为右值引用).
1) There is nothing wrong with passing a reference to a reference (it is what the move-constructor and move-assignment operators use - though it is actually called a rvalue-reference).
2) 你所做的不是将引用传递给引用,而是将相同的引用通过 f
传递给 g
:
2) What you are doing is not passing a reference to a reference, but rather passing the same reference through f
to g
:
void g(int& x)
{
x = 5;
}
void f(int& x)
{
std::cout << "f-in " << x << std::endl;
g(x);
std::cout << "f-out " << x << std::endl;
}
int main()
{
int x = 42;
f(x);
std::cout << "New x = " << x << std::endl;
}
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