当对数组使用auto时，为什么将其转换为指针而不是引用? [英] When auto is used against array, why it's converted to pointer and not reference?

问题描述

请参见以下示例:

int arr[10];
int *p = arr; // 1st valid choice
int (&r)[10] = arr; // 2nd valid choice

现在，当我们对 arr 使用 auto 时，它将选择第一个选择.

Now when we use auto against arr then, it chooses the 1st choice.

auto x = arr; // x is equivalent to *p

是否有必要为数组选择选择指针而不是引用?

Is there a reason for choosing a pointer and not reference for array ?

推荐答案

是.在该表达式中，由于 lvalue-to-rvalue 转换，该数组衰减为指针类型.

Yes. In that expression, the array decays into pointer type, due to lvalue-to-rvalue conversion.

如果要使用 array 类型，而不是 pointer 类型，请执行以下操作:

If you want array type , not pointer type, then do this:

auto & x = arr; //now it doesn't decay into pointer type!

目标类型中的

& 可以防止数组衰减为指针类型！

& in the target type prevents the array from decaying into pointer type!

x 是一个数组，而不是一个指针，可以证明为:

x is an array and not a pointer, can be proved as:

void f(int (&a)[10]) 
{
    std::cout << "f() is called. that means, x is an array!" << std::endl;
}
int main() {
     int arr[10];
     auto & x = arr; 
     f(x); //okay iff x is an int array of size 10, else compilation error!
}

输出:

f() is called. that means, x is an array!

ideone上的演示: http://www.ideone.com/L2Ifp

Demo at ideone : http://www.ideone.com/L2Ifp

请注意，不能使用 pointer 类型调用 f .可以使用大小为 10 的 int 数组来调用它.尝试使用任何其他类型调用它，将导致编译错误.

Note that f cannot be called with pointer type. It can be called with an int array of size 10. Attempting to call it with any other type, will result in compilation error.

这篇关于当对数组使用auto时，为什么将其转换为指针而不是引用?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！