golang的defer执行时机案例分析

package main
import "fmt"func calcFunc(x int, y int) int {return x + y
}func main() {// defer语句的执行顺序是，从右到左,逆序执行deferDemo()// deferDemo1函数demo1 := deferDemo1()fmt.Println(demo1) // 0// deferDemo2函数demo2 := deferDemo2()fmt.Println(demo2) // 1// deferDemo4demo4 := deferDemo4()fmt.Println(demo4) // 1// deferDemo5demo5 := deferDemo5()fmt.Println(demo5) // 1// deferDemo6demo6 := deferDemo6()fmt.Println(demo6) // 2// 难点解析/*分析:defer注册顺序:1.defer fmt.Println("AA", calcFunc(x, calcFunc(x, y)))2.defer fmt.Println("BB", calcFunc(x, calcFunc(x, y)))defer执行顺序:1. fmt.Println("BB", calcFunc(x, calcFunc(x, y)))2. fmt.Println("AA", calcFunc(x, calcFunc(x, y)))*/x := 1y := 2// 注册的时候 x = 1, y = 2 所以执行的时候x = 1, y = 2defer fmt.Println("AA", calcFunc(x, calcFunc(x, y))) // calcFunc(1,1+2)  // 结果是:4x = 10// 注册的时候x = 10, y = 2 所以执行的时候x = 10, y = 2defer fmt.Println("BB", calcFunc(x, calcFunc(x, y))) // calcFunc(10,10+2) // 结果是:22y = 20}

分析：
defer语句的执行顺序是，从右到左,逆序执行
执行deferDemo()函数时，会先执行defer语句，再执行函数体，函数体执行完后，再执行defer语句
执行结果:
开启defer
结束defer
defer语句6
defer语句5
defer语句4
defer语句3
defer语句2
defer语句1

func deferDemo() {fmt.Println("开启defer")defer func() {fmt.Println("defer语句1")}()defer func() {fmt.Println("defer语句2")}()defer func() {fmt.Println("defer语句3")}()defer fmt.Println("defer语句4")defer fmt.Println("defer语句5")defer fmt.Println("defer语句6")fmt.Println("结束defer")
}

deferDemo1执行结果是 0,延迟执行

func deferDemo1() int {var a intdefer func() {a++}()return a
}

deferDemo1执行结果是1：原因返回匿名返回值a等待函数体内操作完成才会执行return

func deferDemo2() (a int) {// 步骤一先赋值//a = 0// 步骤二再执行defer语句defer func() {a++}()// 步骤三再执行函数体// 返回值a等待函数体内操作完成才会执行returnreturn a
}

deferDemo3执行结果是1： 原因返回匿名返回值a等待函数体内操作完成才会执行return

func deferDemo3() (a int) {defer func() {a++}()return a
}

deferDemo4

func deferDemo4() int {x := 1defer func(x int) {x++ // 内部x和外面x不是同一个变量}(x)return x // 1
}

deferDemo5

func deferDemo5() (x int) {x = 1defer func(x int) {x++ // 内部x和外面x不是同一个变量}(x)return x // 1
}

deferDemo6

func deferDemo6() (x int) {x = 1defer func() {x++}()return x // 2
}