深入理解 go slice

看下面的几个例子：

func modifyEle(array []int) {
	for i := 0; i < len(array); i++ {
		(array)[i] = i
	}
	fmt.Printf("modify   int  array len = %d, cap = %d \n", len(array), cap(array))
}

func main() {
	array := []int{4, 3, 2, 1, 0}
	fmt.Println(array)
	
	fmt.Printf("modify befort array len = %d, cap = %d \n", len(array), cap(array))
	modifyEle(array)
	fmt.Printf("modify after  array len = %d, cap = %d \n", len(array), cap(array))
	fmt.Println(array)
}

结果:

[4 3 2 1 0]
modify befort array len = 5, cap = 5 
modify   in   array len = 5, cap = 5 
modify after  array len = 5, cap = 5 
[0 1 2 3 4]

上面这个例子中，slice 可以被修改。 再看一个例子。

func deleteEle(array []int) {
	array = array[:1]

	fmt.Printf("delete   in   array len = %d, cap = %d \n", len(array), cap(array))
}

func main() {
	array := []int{4, 3, 2, 1, 0}
	fmt.Println(array)

	fmt.Printf("delete befort array len = %d, cap = %d \n", len(array), cap(array))
	deleteEle(array)
	fmt.Printf("delete after array len = %d, cap = %d \n", len(array), cap(array))
	fmt.Println(array)
}

结果

[4 3 2 1 0]
delete befort array len = 5, cap = 5 
delete   in   array len = 1, cap = 5 
delete after  array len = 5, cap = 5 
[4 3 2 1 0]

在上面这个例子中，删除slice 里的元素好像没成功。 继续看下面的例子

func modifyEle(array []int) {
	array = append(array, 1)
	for i := 0; i < len(array); i++ {
		(array)[i] = i
	}
	fmt.Printf("modify   in   array len = %d, cap = %d \n", len(array), cap(array))
}

func main() {
	array := []int{4, 3, 2, 1, 0}
	fmt.Println(array)
	fmt.Printf("modify befort array len = %d, cap = %d \n", len(array), cap(array))
	modifyEle(array)
	fmt.Printf("modify after  array len = %d, cap = %d \n", len(array), cap(array))
	fmt.Println(array)
}

这个例子跟第一个很像，但是结果完全不同。

[4 3 2 1 0]
modify befort array len = 5, cap = 5 
modify   in   array len = 6, cap = 10 
modify after  array len = 5, cap = 5 
[4 3 2 1 0]

理解了这三个例子，就完全理解 go slice 了。