英文:
Question about operations that mix numeric types in Go
问题
我知道Go是一种静态类型语言，不允许混合使用数值类型的操作，例如，你不能将int和float64相加：
package main
import (
"fmt"
)
func main() {
var a int = 1
var b float64 = 1.1
fmt.Println(a + b)
}
运行这个程序会导致错误：
invalid operation: a + b (mismatched types int and float64)
但是当我执行1 + 1.1这样的数学运算时，没有声明变量，程序会返回期望的结果2.1：
package main
import (
"fmt"
)
func main() {
fmt.Println(1 + 1.1)
}
所以我的问题是：为什么1 + 1.1可以工作？在直接使用它们进行加法运算时，1和1.1的数值类型是什么？
英文:
I know that Go is a statically typed language that doesn't allow operations that mix numeric types, for example, you can't add an int to a float64:
package main
import (
"fmt"
)
func main() {
var a int = 1
var b float64 = 1.1
fmt.Println(a + b)
}
Running this program will cause an error:
> invalid operation: a + b (mismatched types int and float64)
But when I do the math 1 + 1.1 without declaring the variables, the program returns the desired result which is 2.1:
package main
import (
"fmt"
)
func main() {
fmt.Println(1 + 1.1)
}
So my question is: Why does 1 + 1.1 work? What's the numeric type of 1 and 1.1 when I use them directly in the addition?
答案1
得分: 0
如@Volker所提到的，这是因为1 + 1.1被评估为一个无类型的常量表达式。
因此，下面的程序也可以工作，因为现在a和b都是无类型的常量。
package main
import (
"fmt"
)
func main() {
const a = 1
const b = 1.1
fmt.Println(a + b)
}
英文:
As mentioned by @Volker, this is because 1 + 1.1 is evaluated as an untyped constant expression.
So below program also works because now a and b are both untyped constant.
package main
import (
"fmt"
)
func main() {
const a = 1
const b = 1.1
fmt.Println(a + b)
}
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