英文:
How to calculate an int value and an float64 value in Golang?
问题
在上面的代码中，我无法理解为什么声明 a = int(b) 没有将 b 的值转换为 int。此外，如果我们可以完全省略这行代码并获得相同的结果，那么这样做的意义是什么？
package main
import "fmt"
func main() {
a, b := 10, 5.5
a = int(b)
fmt.Println(float64(a) + b)
}
非常感谢！
英文:
On the following code I could not understand why the declaration a = int(b) didn't convert the value of b to int. Also, what is the point in doing so if we can achieve the same result omitting this line entirely anyways?
package main
import "fmt"
func main() {
a, b := 10, 5.5
a = int(b)
fmt.Println(float64(a) + b)
}
Thanks a lot!
答案1
得分: 1
a, b := 10, 5.5 这行代码中，将值10赋给了变量a，默认类型为int，将值5.5赋给了变量b，默认类型为float64。点击这里查看详情
然后 a = int(b) 这行代码将变量b的整数值赋给了a，即5。如果你不需要保留a的先前值，那么这两行代码可以简化为以下形式，得到相同的结果：
package main
import (
"fmt"
)
func main() {
b := 5.5
a := int(b)
fmt.Println(float64(a) + b) //输出：10.5
}
fmt.Println(float64(a) + b) 你不能对不同类型的值进行运算，所以你需要将a转换为float64类型，然后进行运算。如果将结果赋给另一个变量，那么结果也将是float64类型。
英文:
a, b := 10, 5.5
In this line, value 10 is assigned to the a with default type int and value 5.5 is assigned to the b with default type float64. tour here
then a = int(b) this line get integer value of b, that is 5 and assigned to a. If you don't need a's previous value then those two line can be write as follow and get same result.
package main
import (
"fmt"
)
func main() {
b := 5.5
a := int(b)
fmt.Println(float64(a) + b) //Output: 10.5
}
fmt.Println(float64(a) + b) you can not do operations with different types. that's why you need to convert a to float64 and do the operation. result also a float64 if you assigned it to another variable.
答案2
得分: 0
使用：
a = int(b)
a 变成了 5。所以：
float64(a) + b
变成了 10.5，因为 a=5 且 b=5.5。
英文:
With:
a=int(b)
a becomes 5. So:
float64(a)+b
becomes 10.5, because a=5 and b=5.5
答案3
得分: -1
package main
import "fmt"
func main() {
a, b := 10, 5.5
a = int(b)
fmt.Println(float64(a) + b, a)
}
输出
10.5 5
看起来它将b转换为int -> 5 -> a = a + b = 5 + 5.5 = 10.5
这里将b强制转换为int，并将该值放入a中。
如果你想改变b的值，你应该这样做 b = int(b)。
英文:
package main
import "fmt"
func main() {
a, b := 10, 5.5
a = int(b)
fmt.Println(float64(a) + b, a)
}
Outputs
10.5 5
It seems like it is converting b to int -> 5 -> a = a+b = 5 + 5.5 = 10.5
Here b is typecasted to int and putting that value in a.
If you need to change the value of b you should do b = int(b).
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