# command-line-arguments
../345/7306/4fm0/index.go:12: syntax error: unexpected ), expecting semicolon or newline or }
../345/7306/4fm0/index.go:12: syntax error: unexpected ), expecting semicolon or newline or }
2016-03-30
mm2["scala"]=25
mm2["erlang"]=50
mm2["python"]=0
這樣寫貌似也可以通過啊,,,如果map中已經(jīng)存在,相當(dāng)于直接覆蓋原來的對吧
mm2["erlang"]=50
mm2["python"]=0
這樣寫貌似也可以通過啊,,,如果map中已經(jīng)存在,相當(dāng)于直接覆蓋原來的對吧
2016-03-20
go func() {
var sender Sender = myChannel
sender <- number
}()
go func() {
var receiver Receiver = myChannel
fmt.Println("Received!", <-receiver)
fmt.Println("Sent!")
}()
var sender Sender = myChannel
sender <- number
}()
go func() {
var receiver Receiver = myChannel
fmt.Println("Received!", <-receiver)
fmt.Println("Sent!")
}()
2016-03-06
這道題的判題程序有問題?
代碼應(yīng)該是沒錯吧,當(dāng)通道長為0時,執(zhí)行到sender <- number時第一個go語句塊就被阻塞了,直到通道值被接收,也就是說第二個go語句塊執(zhí)行完,才執(zhí)行fmt.Println("Sent!"),因此最后的結(jié)果Received!應(yīng)該在Sent!之前。
這樣理解對嗎?
代碼應(yīng)該是沒錯吧,當(dāng)通道長為0時,執(zhí)行到sender <- number時第一個go語句塊就被阻塞了,直到通道值被接收,也就是說第二個go語句塊執(zhí)行完,才執(zhí)行fmt.Println("Sent!"),因此最后的結(jié)果Received!應(yīng)該在Sent!之前。
這樣理解對嗎?
2016-03-01