英文:
golang does not update array in a map
问题
m := map[int][2]int{1:{0,10}}
m[1][0] = 1
我期望上面的代码像这样工作：
a := [2]int{0,10}
a[0] = 1
但实际上它会报错 cannot assign to m[1][0]。
可能的解释是什么？
附注：我知道可以通过声明一个 int 到切片的映射来解决这个问题。
英文:
m := map[int][2]int{1:{0,10}}
m[1][0] = 1
I expect the above to work like this
a := [2]int{0,10}
a[0] = 1
but instead it gives the following error cannot assign to m[1][0]
What could be a possible explanation for this?
P.S. I know I can get around the problem by declaring a map of int to slice instead of int to array.
答案1
得分: 4
[翻译结果]
赋值的左操作数必须是可寻址的，一个映射索引表达式或空白标识符。
值m[1][0]是不可寻址的。请参阅规范以获取可寻址的内容列表。映射值不在该列表中。
表达式m[1][0]是一个数组索引表达式，而不是映射索引表达式。
要更新映射中的值，请将映射值赋给一个变量（该变量是可寻址的），更新变量并重新赋值给映射：
t := m[1]
t[0] = 1
m[1] = t
另一种方法是使用指向[2]int的指针映射：
m := map[int]*[2]int{1: {0, 10}}
m[1][0] = 1
由于隐式指针间接引用，数组元素是可寻址的。
英文:
The left-hand operand of an assignment must be addressable, a map index expression or the blank identifier.
The value m[1][0] is not not addressable. See the specification for a list of what is addressable. A map value is not in that list.
The expression m[1][0] is an array index expression, not a map index expression.
To update the value in the map, assign the map value to a variable (which is addressable), update the variable and assign back to the map:
t := m[1]
t[0] = 1
m[1] = t
Another approach is to use a map of pointers to [2]int:
m := map[int]*[2]int{1: {0, 10}}
m[1][0] = 1
The array elements are addressable because of the implicit pointer indirection.
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