问题描述
Go具有非常整齐的多个返回值范例。但是看起来 v，好的：= map [key] 和 v，k：=范围m 具有相同的符号。这是一个简单的示例：
Go has very neat multiple return values paradigm. But It looks like v, ok := map[key] and v, k := range m use different mechanism with same notation. Here is a simple example:
func f2() (k, v string) {
return "Hello", "World"
}
func main(){
k := f2() // Doesn't work : multiple-value f2() in single-value context
m := map[string]int{"One": 1}
// It works
v, ok := m["One"]
// How it all work?
v := m["One"]
for k := range m {}
}
在上面的示例中， k：= f2（）给出错误，因为 f2 返回两个值，而 v，确定：= m [一个] 和 v：= m [一个] -这两个表达式都可以正常工作。
为什么会有不同的行为？
In above example, k := f2() gives error as f2 returns two values, whereas v, ok := m["One"] and v := m["One"] - both expressions work without any error.Why is that different behavior?
推荐答案
从内置的映射中提取，在地图，数组或切片上使用 range ，还使用 type断言允许一个或两个变量。用户定义的函数和方法不是这种情况。如果一个函数声明了两个返回值，则必须告诉它们如何处理这两个值，或者忽略这两个值：
A fetch from the built in map, using range on a map, array or slice, and also type assertions allows for one or two variables. This is not the case for user defined functions and methods. If a function declares two return values, you must tell what to do with both of them, or ignore both:
k, _ := f2() // Specify what to do with each returned value
f2() // Ignoring both
为什么？因为规范说的是这样的：
Why? Because the specification says it is so:
Map (indexed expressions):
v，确定= a [x]
v，确定：= a [x ]
var v，确定= a [x]
v, ok = a[x]
v, ok := a[x]
var v, ok = a[x]
其中索引表达式的结果是一对类型为（V，bool ）。以这种形式，如果键x存在于映射中，则ok的值为true，否则为false。 v的值是单结果形式的值a [x]。
where the result of the index expression is a pair of values with types (V, bool). In this form, the value of ok is true if the key x is present in the map, and false otherwise. The value of v is the value a[x] as in the single-result form.
Range (for statement):
范围表达式：m map [K] V
第一个值：键k K
第二个值（如果存在第二个变量）：m [k] V
Range expression: m map[K]V
1st value: key k K
2nd value (if 2nd variable is present): m[k] V
Type assertion:
和
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