Go语言,语法错误:意外的++,期望:

huangapple go评论83阅读模式
英文:

Go golang, syntax error: unexpected ++, expecting :

问题

func test(args ...string) {
    var msg map[string]interface{}

    i := 0
    msg["product"] = args[i]
    i++
    msg["key"] = args[i]
    i++
    msg["signature"] = args[i]
    i++
    msg["string_to_sign"] = args[i]
}

编译后,我得到以下错误信息:

./utils.go:28: 语法错误:意外的++,期望:
./utils.go:28: 标签后缺少语句
./utils.go:29: 语法错误:意外的++,期望:
./utils.go:30: 语法错误:意外的++,期望:
./utils.go:31: 语法错误:意外的++,期望:
./utils.go:36: 语法错误:意外的++,期望:
./utils.go:37: 语法错误:意外的++,期望:

为什么我不能在切片的索引中使用i++?在切片的索引中有什么限制吗?

英文:
  func test(args ...string) {
    var msg map[string] interface{}

    i := 0
    msg["product"] = args[i++]
    msg["key"] = args[i++]
    msg["signature"] = args[i++]
    msg["string_to_sign"] = args[i++]
  }

  go build utils.go

after compile, I get the error message

    ./utils.go:28: syntax error: unexpected ++, expecting :
    ./utils.go:28: missing statement after label
    ./utils.go:29: syntax error: unexpected ++, expecting :
    ./utils.go:30: syntax error: unexpected ++, expecting :
    ./utils.go:31: syntax error: unexpected ++, expecting :
    ./utils.go:36: syntax error: unexpected ++, expecting :
    ./utils.go:37: syntax error: unexpected ++, expecting :

why can't I put i++ in index of slice? is there any limitation in index of slice?

答案1

得分: 70

> Go常见问题(FAQ)
>
> 为什么++和--是语句而不是表达式?为什么是后缀而不是前缀?
>
> 没有指针算术,前缀和后缀递增运算符的方便性降低了。通过将它们从表达式层次结构中完全移除,表达式语法得到简化,同时也消除了关于++和--的求值顺序的混乱问题(考虑f(i++)和p[i] = q[++i])。这种简化是显著的。至于后缀和前缀,两者都可以正常工作,但后缀版本更加传统;对前缀的坚持起源于STL,这是一个名字中讽刺地包含后缀递增的语言的库。
>
> Go编程语言规范
>
> 递增和递减语句
>
> “++”和“--”语句通过未类型化的常量1递增或递减它们的操作数。与赋值一样,操作数必须是可寻址的或映射索引表达式。
>
> IncDecStmt = Expression ( "++" | "--" ) .
>
> 以下赋值语句在语义上是等价的:
>
> 递增递减语句 赋值语句
> x++ x += 1
> x-- x -= 1

写成:

func test(args ...string) {
	var msg map[string]interface{}
	i := 0
	msg["product"] = args[i]
	i++
	msg["key"] = args[i]
	i++
	msg["signature"] = args[i]
	i++
	msg["string_to_sign"] = args[i]
}

在你的特定情况下,简化为:

func test(args ...string) {
	var msg map[string]interface{}
	msg["product"] = args[0]
	msg["key"] = args[1]
	msg["signature"] = args[2]
	msg["string_to_sign"] = args[3]
}
英文:

> Go Frequently Asked Questions (FAQ)
>
> Why are ++ and -- statements and not expressions? And why postfix,
> not prefix?

>
> Without pointer arithmetic, the convenience value of pre- and postfix
> increment operators drops. By removing them from the expression
> hierarchy altogether, expression syntax is simplified and the messy
> issues around order of evaluation of ++ and -- (consider f(i++) and
> p[i] = q[++i]) are eliminated as well. The simplification is
> significant. As for postfix vs. prefix, either would work fine but the
> postfix version is more traditional; insistence on prefix arose with
> the STL, a library for a language whose name contains, ironically, a
> postfix increment.
>
> The Go Programming Language Specification
>
> IncDec statements
>
> The "++" and "--" statements increment or decrement their operands by
> the untyped constant 1. As with an assignment, the operand must be
> addressable or a map index expression.
>
> IncDecStmt = Expression ( "++" | "--" ) .
>
> The following assignment statements are semantically equivalent:
>
> IncDec statement Assignment
> x++ x += 1
> x-- x -= 1

Write,

func test(args ...string) {
	var msg map[string]interface{}
	i := 0
	msg["product"] = args[i]
	i++
	msg["key"] = args[i]
	i++
	msg["signature"] = args[i]
	i++
	msg["string_to_sign"] = args[i]
}

Which, in your particular case, simplifies to,

func test(args ...string) {
	var msg map[string]interface{}
	msg["product"] = args[0]
	msg["key"] = args[1]
	msg["signature"] = args[2]
	msg["string_to_sign"] = args[3]
}

答案2

得分: 20

根据语言规范,http://golang.org/ref/spec#IncDec_statements,i++ 是一个“IncDec语句”,它是一个“语句”,而不是一个“表达式”。至于 args[index],index 必须是一个“表达式”。如果你想了解更多细节,只需阅读Go语言规范,这正是语言要求的。

英文:

According to Language Specification, http://golang.org/ref/spec#IncDec_statements, i++ is a IncDec statements, which is a statement, but not a expression.As for args[index], index must be a expression. You want more details , just read it Go Language Specification, it's just what the language demand.

答案3

得分: 14

正如其他人所说,i++ 在 Go 中是一个语句,而不是像在 C 中那样是一个表达式。Go 有一种不同的方式来表达相同的意图,即使用多重赋值:

func test(args ...string) {
    msg := make(map[string]string)
    i := 0

    msg["product"], i = args[i], i+1
    msg["key"], i = args[i], i+1
    msg["signature"], i = args[i], i+1
    msg["string_to_sign"], i = args[i], i+1

    fmt.Printf("%v\n", msg)
}

你对 map 的定义在运行时也会失败。

英文:

As other people have said i++ is a statement in go, not an expression as it is in C. Go has a different way of expressing the same intent using multiple assignment:

func test(args ...string) {
    msg := make(map[string]string)
    i := 0

    msg["product"], i = args[i], i+1
    msg["key"], i = args[i], i+1
    msg["signature"], i = args[i], i+1
    msg["string_to_sign"], i = args[i], i+1

    fmt.Printf("%v\n", msg)
}

Your definition of map would have failed at runtime too.

答案4

得分: 5

++ 运算符令人失望。这是我的修改:

func test(args ...string) {
    i := 0
    inc := func(i *int) int { *i++; return *i }
    var msg map[string]interface{}

    msg["product"] = args[inc(&i)]
    msg["key"] = args[inc(&i)]
    msg["signature"] = args[inc(&i)]
    msg["string_to_sign"] = args[inc(&i)]
}
英文:

++ operator is disappointing. This is my hack:

func test(args ...string) {
    i := 0
    inc := func(i *int) int { *i++; return i }
    var msg map[string] interface{}

    msg["product"] = args[inc(&i)]
    msg["key"] = args[inc(&i)]
    msg["signature"] = args[inc(&i)]
    msg["string_to_sign"] = args[inc(&i)]
  }

huangapple
  • 本文由 发表于 2014年9月12日 11:24:33
  • 转载请务必保留本文链接:https://go.coder-hub.com/25800242.html
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