我最近開(kāi)始學(xué)習(xí) Go，我寫過(guò)一個(gè)案例，我無(wú)法理解為什么他會(huì)根據(jù)僅打印 [第 13 行] 的一行中所做的更改獲得兩種不同的行為，在第一次運(yùn)行中，我使用 [第 13 行] 運(yùn)行程序，然后在主例程中，當(dāng)我在 [第 21 行] 打印通道長(zhǎng)度時(shí)打印 0，在下一行打印 2 之后（我'談?wù)撝饕绦蛑谱鞯牡谝粋€(gè)印刷品）。在第二次運(yùn)行中，我刪除了 [第 13 行]，然后在第一次打印時(shí)，通道的長(zhǎng)度為 2。在圖片中，您可以在控制臺(tái)中看到兩個(gè)不同的打印件，我不明白為什么它們不同 [只是因?yàn)樘砑?刪除第 13 行]。// Go behavior that I do not understand /:package mainimport "fmt"func main() {    mychnl := make(chan string, 2)    go func(input chan string) {        input <- "Inp1"        // If we remove this line the length of the chan in the print will be equal in both prints        fmt.Println("Tell me why D:")        input <- "Inp2"        input <- "Inp3"        input <- "Inp4"        close(input)    }(mychnl)    for res := range mychnl {        fmt.Printf("Length of the channel is: %v The received value is: %s length need to be == ", len(mychnl), res)        fmt.Println(len(mychnl))    }}/*Output ->Line 13 = fmt.Println("Tell me why D:")First run with line 13:Tell me why D:Length of the channel is: 0 The received value is: Inp1 length need to be == 2Length of the channel is: 2 The received value is: Inp2 length need to be == 2Length of the channel is: 1 The received value is: Inp3 length need to be == 1Length of the channel is: 0 The received value is: Inp4 length need to be == 0Second run without line 13:Length of the channel is: 2 The received value is: Inp1 length need to be == 2Length of the channel is: 2 The received value is: Inp2 length need to be == 2Length of the channel is: 1 The received value is: Inp3 length need to be == 1Length of the channel is: 0 The received value is: Inp4 length need to be == 0*/