ios - swift中inout参数传入一个函数，在函数体内print怎么没有被执行？

Question

在一个交换函数中，a和b的值进行交换，使用inout参数传值，这样函数对参数所做的修改将会影响参数本身，但是为什么在swap函数里的print没有被执行？

func swap(inout a : Int , inout b : Int){
    let tmp = a
    print("123")
    a = b
    b = tmp
    
}

var a = 6
var b = 9
print("交换之前，a的值是\(a),b的值是\(b)")
swap(&a, &b)
print("交换之后，a的值是\(a),b的值是\(b)")

输出的结果是：
交换之前，a的值是6,b的值是9
交换之后，a的值是9,b的值是6

swap函数里的print哪里去了？？？

Answer 1

I was confused when I first read the question.
I opened a Playground and tried it, only to find out:

The swap function you called is obviously provided by Swift itself!

Although you defined a new swap function above.

So, the solution is to change the function name...

---

Let me tell you by the way:
In Swift, you can exchange two values ​​​​like this

(a, b) = (b, a)