英文:
How to perform division in Go
问题
我正在尝试在Go语言中进行简单的除法运算。
fmt.Println(3/10)
这段代码输出的结果是0，而不是0.3。这有点奇怪。请问有人可以分享一下这背后的原因吗？我还想在Go语言中进行其他算术运算。
谢谢。
英文:
I am trying to perform a simple division in Go.
fmt.Println(3/10)
This prints 0 instead of 0.3. This is kind of weird. Could someone please share what is the reason behind this? i want to perform different arithmetic operations in Go.
Thanks
答案1
得分: 121
二元操作 3 / 10 的操作数是未类型化的常量。根据规范 关于带有未类型化常量的二元操作的说明：
> 如果二元操作的操作数是不同类型的未类型化常量，操作和（对于非布尔操作）结果使用此列表中后面出现的类型：整数、符文、浮点数、复数。
因为 3 和 10 都是未类型化的整数常量，所以表达式的值是未类型化的整数（在这种情况下为 0）。
要获得浮点数常量的结果，其中一个操作数必须是浮点数常量。以下表达式的结果是未类型化的浮点数常量 0.3：
3.0 / 10.0
3.0 / 10
3 / 10.0
当除法操作具有未类型化常量操作数和类型化操作数时，类型化操作数决定表达式的类型。确保类型化操作数是 float64，以获得 float64 的结果。
以下表达式将 int 变量转换为 float64，以获得 float64 结果 0.3：
var i3 = 3
var i10 = 10
fmt.Println(float64(i3) / 10)
fmt.Println(3 / float64(i10))
英文:
The operands of the binary operation 3 / 10 are untyped constants. The specification says this about binary operations with untyped constants
> if the operands of a binary operation are different kinds of untyped constants, the operation and, for non-boolean operations, the result use the kind that appears later in this list: integer, rune, floating-point, complex.
Because 3 and 10 are untyped integer constants, the value of the expression is an untyped integer (0 in this case).
To get a floating-point constant result, one of the operands must be a floating-point constant. The following expressions evaluate to the untyped floating-point constant 0.3:
3.0 / 10.0
3.0 / 10
3 / 10.0
When the division operation has an untyped constant operand and a typed operand, the typed operand determines the type of the expression. Ensure that the typed operand is a float64 to get a float64 result.
The expressions below convert int variables to a float64 to get the float64 result 0.3:
var i3 = 3
var i10 = 10
fmt.Println(float64(i3) / 10)
fmt.Println(3 / float64(i10))
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