而函数的返回值只会在运行计算,所以以上程序会报错const initializer math.Sqrt(5) is not a constant 常量类型 常量可以规定类型,也可以不规定类型。 如果不规定类型,那么它的类型就是不确定的。 例: packagemainimport("fmt")funcmain(){consta,b=3,4//赋值常量 a,b ,ab没有类型varcfloat64=a//定义一...
函数math.Sqrt(4)只会在运行的时候计算,因此const b = math.Sqrt(4)将会抛出错误error main.go:11: const initializer math.Sqrt(4) is not a constant) 字符串常量 双引号中的任何值都是 Go 中的字符串常量。例如像Hello World或Sam等字符串在 Go 中都是常量。 什么类型的字符串属于常量?答案是无类型的...
packagemainfuncmain(){conste=[1]int{1}} Output constinitializer[1]intliteral is not a constant
packagemainfuncmain(){conste=map[string]int{"a":1,}} Output constinitializermap[string]intliteral is not a constant
const initializer \[\]string literal is not a constant [Unlike constants in JavaScript](https://qvault.io/2020/10/22/constants-in-go-vs-javascript-and-when-to-use-them/), Go doesn't allow complex types like slices, maps, or arrays to be constant! Our first instinct may be to lazily...
constMask =1<<3// correctconstPath = os.Getenv("PATH")// incorrect : const initializer os.Getenv("PATH") is not a constant 字面常量(literal) 字面常量(literal),是指程序中硬编码的常量。 golang中字面常量是无类型的,只要该字面常量在相应类型的值域范围内,就可作为该类型的常量。
{'a'}) // not a constant: []byte{'a'} is not a constant (*int)(nil) // not a constant: nil is not a constant, *int is not a boolean, numeric, or string type int(1.2) // illegal: 1.2 cannot be represented as an int string(65.0) // illegal: 65.0 is not an integer ...
In other words, Object is a discriminated union of 8 possible types, and we commonly use a type switch to distinguish them. Object = *Func // function, concrete method, or abstract method | *Var // variable, parameter, result, or struct field | *Const // constant | *TypeName // ...
// example.go:11:7: const initializer add(512, 513) is not a constant 1. 2. 3. spec给出的实例是调用的内置函数,内置函数也只有在参数是常量的情况下被调用才算做常量表达式: const ( c4 = len([10]float64{imag(2i)}) // imag(2i) is a constant and no function call is issued ...
If the last token before a newline is an identifier (which includes words like int and float64), a basic literal such as a number or string constant, or one of the tokens break continue fallthrough return ++ -- ) } the lexer always inserts a semicolon after the token. This could be...