golang中,map作为函数参数是如何传递的
当你声明一个map的时候:
m := make(map[int]int)
编译器会调用 runtime.makemap
:
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// makemap implements a Go map creation make(map[k]v, hint)
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// If the compiler has determined that the map or the first bucket
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// can be created on the stack, h and/or bucket may be non-nil.
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// If h != nil, the map can be created directly in h.
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// If bucket != nil, bucket can be used as the first bucket.
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func makemap(t *maptype, hint int64, h *hmap, bucket unsafe.Pointer) *hmap
所以实际上是返回一个hmap的指针。
如何验证呢?
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func main(){
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m := make(map[int]int)
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m[1] = 1
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fmt.Printf("原始map的内存地址是:%p\n", m)
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modify(m)
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fmt.Println("map值被修改了,新值为:", m)
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}
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func modify(m interface{}){
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fmt.Printf("函数里接收到map的内存地址是:%p\n", p)
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m := p.(map[int]int)
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m[1] = 2
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}
输出结果:
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原始map的内存地址是:0xc00009e030
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函数里接收到map的内存地址是:0xc00009e030
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map值被修改了,新值为: map[1:2]
在main函数中,m是个指针变量,它保存的值是:0xc00009e030。
在modify函数中,m也是个指针变量,保存的值也是:0xc00009e030。
说明初始化map后,返回的是指针变量,在函数之间,传递的是map的地址。
map和channel是相似的。
那么为什么不是 *map[key]value呢,这样不是更直观?
Ian Taylor answered this recently in a golang-nuts 原话是:
In the very early days what we call maps now were written as pointers, so you wrote *map[int]int. We moved away from that when we realized that no one ever wrote
map
without writing\*map
.
意思是,一开始也是写作 *map[key]value,后来发现所有的map都是当作指针来用的,于是就省略简写了。
转载于:https://www.cnblogs.com/zhouj-happy/p/10962500.html
posted on 2021-04-01 11:15 ExplorerMan 阅读(430) 评论(0) 编辑 收藏 举报