golang的new函數

In Golang, to allocate memory, we have two built-in functions new() and make().

在Golang中，要分配內存，我們有兩個內置函數new()和make() 。

1)new()函數 (1) new() function)

• Memory returned by new() is zeroed.

  new()返回的內存為零。

• new() only returns pointers to initialized memory.

  new()僅返回指向初始化內存的指針。

• new() works for all the data types (except channel, map), and dynamically allocates space for a variable of that type and initialized it to zero value of that type and return a pointer to it.

  new()適用于所有數據類型(通道，映射除外)，并為該類型的變量動態分配空間，并將其初始化為該類型的零值并返回指向它的指針。

Example:

例：

result = new(int)

is equivalent to

相當于

var temp int // declare an int type variablevar result *int // declare a pointer to intresult = &temp

Example/program:

示例/程序：

There are three different ways to create a pointer that points to a zeroed structure value, each of which is equivalent:

有三種不同的方法可以創建指向零結構值的指針，每種方法都等效：

package mainimport "fmt"type Sum struct {x_val inty_val int }func main() {// Allocate enough memory to store a Sum structure value// and return a pointer to the value's addressvar sum Sump := &sumfmt.Println(p)// Use a composite literal to perform //allocation and return a pointer// to the value's addressp = &Sum{}fmt.Println(p)// Use the new function to perform allocation, //which will return a pointer to the value's address.p = new(Sum)fmt.Println(p) }

Output

輸出量

&{0 0} &{0 0} &{0 0}

2)make()函數 (2) make() function)

make() only makes slices, maps, and channels. make returns value of type T(data type) not *T

Example of slices:

make([]int, 10, 20) – Here, make creates the slice, and initialize its content depending on the default data type value. here int is used, so the default value is 0.

new([20]int)[0:10] – Here, It will also create slice but returns pointers to initialized memory.

Example/program:

There are two different ways to initialize a map which maps string keys to bool values are given below.

package mainimport "fmt"func main() {// Using make() to initialize a map.m := make(map[string]bool, 0)fmt.Println(m)// Using a composite literal to initialize a map.m = map[string]bool{}fmt.Println(m) }

Output

Reference: allocation_new

TOP Interview Coding Problems/Challenges

• Run-length encoding (find/print frequency of letters in a string)

• Sort an array of 0's, 1's and 2's in linear time complexity

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make()僅制作切片，地圖和通道。 make返回類型T (數據類型)的值，而不是* T

切片示例：

make([] int，10，20) –在這里，make創建切片，并根據默認數據類型值初始化其內容。 這里使用int，所以默認值為0。

new([20] int)[0:10] –在這里，它還將創建切片，但返回指向已初始化內存的指針。

示例/程序：

有兩種不同的初始化映射的方法，將字符串鍵映射到bool值的方法如下。

package mainimport "fmt"func main() {// Using make() to initialize a map.m := make ( map [ string ] bool , 0 )fmt.Println(m)// Using a composite literal to initialize a map.m = map [ string ] bool {}fmt.Println(m) }

輸出量

參考： allocation_new

最佳面試編碼問題/挑戰

• 游程編碼(字符串中字母的查找/打印頻率)

• 以線性時間復雜度對0、1和2的數組進行排序

• 檢查字謎(檢查兩個字符串是否是字謎)

• 相對排序算法

• 查找給定總和的子數組

• 在給定總和K的二叉樹中找到級別

• 檢查二叉樹是否為BST(二叉搜索樹)

• 1 [0] 1個樣式計數

• 大寫一行中每個單詞的第一個和最后一個字母

• 打印二叉樹的垂直和

• 打印二叉樹的邊界和

• 反轉單個鏈表

• 解決主要算法問題的貪婪策略

• 工作排序問題

• 根到葉的路徑總和

• 矩陣中的出口點

• 在鏈表中查找循環長度

• 一流的禮帽

• 打印所有沒有兄弟的節點

• 轉換為求和樹

• 最短的源到目標路徑

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翻譯自: https://www.includehelp.com/golang/new-and-make-functions-with-examples.aspx

golang的new函數

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