问题描述
众所周知，声明指针的语义准确方法是
It's pretty common knowledge that the semantically accurate way to declare pointers is
int *x;
代替
int* x;
这是因为 C 将 *x 视为 int，而不是将 x 视为 int 指针.
This is because C sees *x as an int, not x as an int pointer.
这很容易证明
int* a, b;
其中a 是一个int 指针，而b 是一个int.
where a is an int pointer, while b is an int.
在 Stack Overflow 上至少有 5 个重复的问题讨论了这个问题的指针.但是引用呢?
There are at least 5 duplicate questions on Stack Overflow that discuss this issue for pointers. But what about references?
推荐答案
在研究这个问题时，我已经找到了答案:
While researching for this question, I already found the answer:
& 需要像 * 一样编写.
The & needs to be written just like the *.
演示代码类似于指针演示代码:
The demonstration code is similar to the pointer demonstration code:
int main() {
int a = 0;
int b = 1;
int& ar = a, br = b;
br = 2;
return b;
}
这返回 1，这意味着 ar 是一个 int 引用，而 br 只是一个整数.
This returns 1, which means that ar is an int reference, while br is just an integer.
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