Go-可变参数函数传递

问题描述：

Situation:

I'm trying to write a simple fmt.Fprintf wrapper which takes a variable number of arguments. This is the code:

func Die(format string, args ...interface{}) {
    str := fmt.Sprintf(format, args)
    fmt.Fprintf(os.Stderr, "%v
", str)
    os.Exit(1)
}

Problem:

When I call it with Die("foo"), I get the following output (instead of "foo"):

foo%!(EXTRA []interface {}=[])

Why is there "%!(EXTRA []interface {}=[])" after the "foo"?
What is the correct way to create wrappers around fmt.Fprintf?

情况： strong> p>

我正在尝试编写一个简单的fmt.Fprintf code>包装器，该包装器使用可变数量的参数。这是代码： p>

  func Die（格式字符串，args ... interface {}）{
 str：= fmt.Sprintf（format，args）
 fmt  .Fprintf（os.Stderr，“％v 
”，str）
 os.Exit（1）
} 
  code>  pre> 
 
  问题：  p> 
 
 
当我用 Die（“ foo”） code>调用它时，我得到以下输出（而不是“  foo  em>  “）： p> 
 
 
 
  foo％！（额外[]接口{} = []） p> 
  blockquote> 
 
 
为什么在“  foo  em>”后面有“ ％！（EXTRA [] interface {} = []） em>”？ li> 
 在 fmt.Fprintf  code>周围创建包装的正确方法是什么？ li> 
  ul> 
  div>

答

Variadic functions receive the arguments as a slice of the type. In this case your function receives a []interface{} named args. When you pass that argument to fmt.Sprintf, you are passing it as a single argument of type []interface{}. What you really want is to pass each value in args as a separate argument (the same way you received them). To do this you must use the ... syntax.

str := fmt.Sprintf(format, args...)

This is also explained in the Go specification here.

Go-可变参数函数传递

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