package main
import "fmt"
func main() {
var numbers2 [5]int
numbers2[0] = 2
numbers2[2]=0
numbers2[3] = numbers2[0] - 3
numbers2[1] = numbers2[2] + 5
numbers2[4] = len(numbers2)
sum := (2+5+0-1+5)
fmt.Printf("%v\n", (sum == numbers2[0]+numbers2[1]+numbers2[2]+numbers2[3]+numbers2[4]))
}
import "fmt"
func main() {
var numbers2 [5]int
numbers2[0] = 2
numbers2[2]=0
numbers2[3] = numbers2[0] - 3
numbers2[1] = numbers2[2] + 5
numbers2[4] = len(numbers2)
sum := (2+5+0-1+5)
fmt.Printf("%v\n", (sum == numbers2[0]+numbers2[1]+numbers2[2]+numbers2[3]+numbers2[4]))
}
2017-08-04
fmt.Printf("浮點(diǎn)數(shù) %E 表示的是 %f。\n", num2, (num2))
fmt.Printf("浮點(diǎn)數(shù) %E 表示的是 %g\n", num2, (num2))
結(jié)果一樣 哈
%g 用最少的數(shù)字來(lái)表示
%f (=%.6f) 6位小數(shù)點(diǎn)
fmt.Printf("浮點(diǎn)數(shù) %E 表示的是 %g\n", num2, (num2))
結(jié)果一樣 哈
%g 用最少的數(shù)字來(lái)表示
%f (=%.6f) 6位小數(shù)點(diǎn)
2017-08-03
輸出結(jié)果為: 類型為 uint64 的整數(shù) 65535 需占用的存儲(chǔ)空間為 8 個(gè)字節(jié)。
2017-08-03
看了各位的評(píng)論,個(gè)人覺(jué)得可能是算法問(wèn)題??赡芨饔兴L(zhǎng)吧
2017-07-25
func (p *Person) Move(newaddr string) string{
oldaddr := p.Address
p.Address = newaddr
return oldaddr
}
oldaddr := p.Address
p.Address = newaddr
return oldaddr
}
2017-07-16
vim /etc/profile
最后一行插入
export GOROOT=/usr/local/go
export GOPATH=~/golib:~/goproject
export GOBIN=~/gobin
export PATH=$PATH:$GOROOT/bin:$GOBIN
然后 source /etc/profile
root 權(quán)限下,
最后一行插入
export GOROOT=/usr/local/go
export GOPATH=~/golib:~/goproject
export GOBIN=~/gobin
export PATH=$PATH:$GOROOT/bin:$GOBIN
然后 source /etc/profile
root 權(quán)限下,
2017-07-12
number=6,然后sender接收到6,但是如果聲明myChannel=make(chan int, (0)),這表示是一個(gè)非緩沖通道,那么,sender為6,myChannel接收到值之后立馬阻塞,然后去執(zhí)行下面的匿名函數(shù),在打印的時(shí)候取走了myChannel通道里面的值,然后myChannel不被阻塞,再執(zhí)行上面的一個(gè)匿名函數(shù),所以結(jié)果是Received! 6
Sent!
Sent!
2017-07-11