英文:
How does multiple variable assignment work?
问题
请检查下面的示例代码，并查看第三行。
a := [3]int{10,20}
var i int = 50
i, a[2] = 100, i
fmt.Println(i) //100
fmt.Println(a) //[10 20 50]
我已经覆盖了变量i中的值100，并立即应用到整型数组中。当我打印数组时，新值没有被打印出来。在Go语言中，多变量赋值是如何工作的？为什么i的值没有立即更新到数组中？
英文:
Please check the below sample code, and look into 3rd line.
a := [3]int{10,20}
var i int = 50
i, a[2] = 100, i
fmt.Println(i) //100
fmt.Println(a) //[10 20 50]
I have overwritten the value 100 in i variable and immediately applied the int array. When I printed the array, the new value was not printed. How does multiple variable assignment work in Go? Why the i value is not updated into the array immediately?
答案1
得分: 4
Go语言规范中的赋值部分提到：
> 赋值过程分为两个阶段。
> - 首先，左边的索引表达式和指针间接引用（包括选择器中的隐式指针间接引用）以及右边的表达式都按照通常的顺序进行求值。
- 其次，按照从左到右的顺序进行赋值。
这意味着：
var i int = 50
i, a[2] = 100, i
-
a[2]被赋值为在赋值之前求值的i（即50） -
i被赋值为100
英文:
The assigment section of the Go spec mentions:
> The assignment proceeds in two phases.
> - First, the operands of index expressions and pointer indirections (including implicit pointer indirections in selectors) on the left and the expressions on the right are all evaluated in the usual order.
- Second, the assignments are carried out in left-to-right order.
That means:
var i int = 50
i, a[2] = 100, i
-
a[2]is assigned theievaluated before assignment (50) -
iis assigned100
答案2
得分: 3
这是在Go语言规范中描述和解释的。
基本上，这是一个语句，它同时将两个值分配给两个变量。语句的效果在语句完全执行时可见，就像其他任何表达式一样。
变量i的值在第4行被执行时发生改变，所以在将值赋给a[3]时，它的值仍然是50。
英文:
This is intended and described in the Go language specs.
Basically it is one statement which happens to assign 2 values to 2 variables. The effects of the statement are available/visible when the statement is fully executed, like with any other expression.
The value of i changes the moment you "hit" line 4, so at the time of the assignment to a[3] its value is still 50.
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