迹忆客 计算机编程题库

题库 > Go > Go 笔试题精选 练习:14

Go 笔试题精选

下面代码运行时输出的结果是什么? ```go package main import ( "fmt" "time" ) func main() { ch1 := make(chan int) go fmt.Println(<-ch1) ch1 <- 5 time.Sleep(1 * time.Second) } ```
  • 5
  • 编译不通过
  • 运行时死锁
正确答案是:C
正确率:57%

解析:

在 Go 语言规范中,关于 go 语句有这么一句描述:

GoStmt = "go" Expression .

The expression must be a function or method call; it cannot be parenthesized. Calls of built-in functions are restricted as for expression statements[2].

The function value and parameters are evaluated as usual[3] in the calling goroutine, but unlike with a regular call, program execution does not wait for the invoked function to complete.

这里说明,go 语句后面的函数调用,其参数会先求值,这和普通的函数调用求值一样。在规范中调用部分是这样描述的:

Given an expression f of function type F,

f(a1, a2, … an)

calls f with arguments a1, a2, … an. Except for one special case, arguments must be single-valued expressions assignable[5] to the parameter types of F and are evaluated before the function is called.

大意思是说,函数调用之前,实参就被求值好了。

因此这道题目 go fmt.Println(<-ch1) 语句中的 <-ch1 是在 main goroutine 中求值的。这相当于一个无缓冲的 chan,发送和接收操作都在一个 goroutine 中(main goroutine)进行,因此造成死锁。

关于通道可以参考 Go Channel 缓冲详细介绍

查看笔记

扫码一下
查看教程更方便