在 Go 中,如何保持 for 循环,但如果用户猜测太高或太低则打印不同的输出
In Go, how do stay in a for loop, but print a different output if user guesses too high or too low
我正在尝试在 GO 中开发一个猜谜游戏,让用户尝试 3 次以正确猜出随机数。
它似乎大部分都有效,我正在使用 for 循环来计算剩余的生命数。如果我猜对了数字,它会打印出我想要的:
"You guessed it!"
但问题是,例如,随机数是5。如果我猜 4,它会打印
"Too low"
这是正确的,但在第二次尝试中,如果我要猜测 6,它仍然会说
"Too low"
或者,如果随机数是 5,并且在第一次尝试时我猜到了 6,它会打印
"Too high"
如果我第二次猜 4,它仍然会说
"Too high"
我想我在某个地方陷入了 for 循环,但我不知道如何在不将尝试计数重置回 3 的情况下退出它。
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
secret := getRandomNumber()
fmt.Println(secret)
fmt.Println("Guess the number")
_guess := guessRandomNumber()
for i := 3; i > 0; i-- {
if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i, "guesses left.")
} else if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
} else if _guess > secret {
fmt.Println("Too high")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
}
if _guess == secret {
fmt.Println("You guessed it!")
i = 0
} else if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
} else if _guess > secret {
fmt.Println("Too high")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
}
}
}
func getRandomNumber() int {
rand.Seed(time.Now().UnixNano())
return rand.Int() % 11
}
func guessRandomNumber() int {
var guess int
//fmt.Println("Guess again:")
fmt.Scanf("%d", &guess)
return guess
}
如有任何帮助,我们将不胜感激。到目前为止,我真的很喜欢 GO。
您没有将函数 guessRandomNumber()
的输出分配给 _guess
。除此之外,您还使用了额外的 if-else
,这不是必需的。此外,您还向用户提供了 4 次尝试,而不是 3 次。您可以尝试:
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
secret := getRandomNumber()
fmt.Println(secret)
fmt.Println("Guess the number")
var _guess int
for i := 3; i > 0; i-- {
_guess = guessRandomNumber()
if _guess == secret {
fmt.Println("You guessed it!")
break
} else if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i-1, "guesses left.")
} else {
fmt.Println("Too high")
fmt.Println("You have", i-1, "guesses left.")
}
if i != 1 {
fmt.Println("Guess Again:")
}
}
}
func getRandomNumber() int {
rand.Seed(time.Now().UnixNano())
return rand.Int() % 11
}
func guessRandomNumber() int {
var guess int
//fmt.Println("Guess Number:")
fmt.Scanf("%d", &guess)
return guess
}
我最终使用带有 switch 语句的 for 循环来解决问题。
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
secret := getRandomNumber()
fmt.Println(secret)
fmt.Println("Guess the number")
_guess := guessRandomNumber()
guessCorrect := false
guessCount := 3
for guessCorrect == false && guessCount != 0 {
switch {
case _guess < secret:
fmt.Println("Too low,", guessCount, "guesses left.")
fmt.Scanf("%d", &_guess)
guessCount--
continue
case _guess > secret:
fmt.Println("Too high", guessCount, "guesses left.")
fmt.Scanf("%d", &_guess)
guessCount--
continue
default:
fmt.Println("You got it!")
guessCorrect = true
}
}
}
func getRandomNumber() int {
rand.Seed(time.Now().UnixNano())
return rand.Int() % 11
}
func guessRandomNumber() int {
var guess int
fmt.Scanf("%d", &guess)
return guess
}
我正在尝试在 GO 中开发一个猜谜游戏,让用户尝试 3 次以正确猜出随机数。
它似乎大部分都有效,我正在使用 for 循环来计算剩余的生命数。如果我猜对了数字,它会打印出我想要的:
"You guessed it!"
但问题是,例如,随机数是5。如果我猜 4,它会打印
"Too low"
这是正确的,但在第二次尝试中,如果我要猜测 6,它仍然会说
"Too low"
或者,如果随机数是 5,并且在第一次尝试时我猜到了 6,它会打印
"Too high"
如果我第二次猜 4,它仍然会说
"Too high"
我想我在某个地方陷入了 for 循环,但我不知道如何在不将尝试计数重置回 3 的情况下退出它。
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
secret := getRandomNumber()
fmt.Println(secret)
fmt.Println("Guess the number")
_guess := guessRandomNumber()
for i := 3; i > 0; i-- {
if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i, "guesses left.")
} else if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
} else if _guess > secret {
fmt.Println("Too high")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
}
if _guess == secret {
fmt.Println("You guessed it!")
i = 0
} else if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
} else if _guess > secret {
fmt.Println("Too high")
fmt.Println("You have", i, "guesses left.")
guessRandomNumber()
}
}
}
func getRandomNumber() int {
rand.Seed(time.Now().UnixNano())
return rand.Int() % 11
}
func guessRandomNumber() int {
var guess int
//fmt.Println("Guess again:")
fmt.Scanf("%d", &guess)
return guess
}
如有任何帮助,我们将不胜感激。到目前为止,我真的很喜欢 GO。
您没有将函数 guessRandomNumber()
的输出分配给 _guess
。除此之外,您还使用了额外的 if-else
,这不是必需的。此外,您还向用户提供了 4 次尝试,而不是 3 次。您可以尝试:
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
secret := getRandomNumber()
fmt.Println(secret)
fmt.Println("Guess the number")
var _guess int
for i := 3; i > 0; i-- {
_guess = guessRandomNumber()
if _guess == secret {
fmt.Println("You guessed it!")
break
} else if _guess < secret {
fmt.Println("Too low")
fmt.Println("You have", i-1, "guesses left.")
} else {
fmt.Println("Too high")
fmt.Println("You have", i-1, "guesses left.")
}
if i != 1 {
fmt.Println("Guess Again:")
}
}
}
func getRandomNumber() int {
rand.Seed(time.Now().UnixNano())
return rand.Int() % 11
}
func guessRandomNumber() int {
var guess int
//fmt.Println("Guess Number:")
fmt.Scanf("%d", &guess)
return guess
}
我最终使用带有 switch 语句的 for 循环来解决问题。
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
secret := getRandomNumber()
fmt.Println(secret)
fmt.Println("Guess the number")
_guess := guessRandomNumber()
guessCorrect := false
guessCount := 3
for guessCorrect == false && guessCount != 0 {
switch {
case _guess < secret:
fmt.Println("Too low,", guessCount, "guesses left.")
fmt.Scanf("%d", &_guess)
guessCount--
continue
case _guess > secret:
fmt.Println("Too high", guessCount, "guesses left.")
fmt.Scanf("%d", &_guess)
guessCount--
continue
default:
fmt.Println("You got it!")
guessCorrect = true
}
}
}
func getRandomNumber() int {
rand.Seed(time.Now().UnixNano())
return rand.Int() % 11
}
func guessRandomNumber() int {
var guess int
fmt.Scanf("%d", &guess)
return guess
}