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Adia

请教一个很基本的 goroutine 问题

  •  
  •   Adia · Jun 11, 2017 · 1087 views
    This topic created in 3242 days ago, the information mentioned may be changed or developed.

    第一段代码:

    package main
    
    import (
    	"fmt"
    	"time"
    )
    
    func main() {
    	names := []string{"Eric", "Harry", "Robert", "Jim", "Mark"}
    	for _, name := range names {
    		go func() {
    			fmt.Printf("Hello, %s!\n", name)
    		}()
    	}
    	time.Sleep(time.Millisecond)
    }
    

    结果:

    package main
    
    import (
    	"fmt"
    	"time"
    )
    
    func main() {
    	names := []string{"Eric", "Harry", "Robert", "Jim", "Mark"}
    	for _, name := range names {
    		go func() {
    			fmt.Printf("Hello, %s!\n", name)
    		}()
    	}
    	time.Sleep(time.Millisecond)
    }
    

    第二段代码:

    package main
    
    import (
    	"fmt"
    	"time"
    )
    
    func main() {
    	names := []string{"Eric", "Harry", "Robert", "Jim", "Mark"}
    	for _, name := range names {
    		go func(who string) {
    			fmt.Printf("Hello, %s!\n", who)
    		}(name)
    	}
    	time.Sleep(time.Millisecond)
    }
    

    结果:

    package main
    
    import (
    	"fmt"
    	"time"
    )
    
    func main() {
    	names := []string{"Eric", "Harry", "Robert", "Jim", "Mark"}
    	for _, name := range names {
    		go func(who string) {
    			fmt.Printf("Hello, %s!\n", who)
    		}(name)
    	}
    	time.Sleep(time.Millisecond)
    }
    
    

    问题

    为什么第二段就能正常的迭代了呢?

    4 replies    2018-01-25 16:29:49 +08:00
    Adia
        1
    Adia  
    OP
       Jun 12, 2017
    up
    CuminLo
        2
    CuminLo  
       Jun 17, 2017
    没看懂?结果是?
    Yoock
        3
    Yoock  
       Jun 18, 2017
    我觉得原因可能是每次 for range k 和 v 不是新建变量,而是变量复用。
    所以每次不指定输出的字符串,并发输出 v 的话,都会变成最后一次 v 的值。
    说的有点乱,希望你可以理解。
    xinQing
        4
    xinQing  
       Jan 25, 2018
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