英文:
Why i cant apply address operator (&) to functions return values directly
问题
我不明白为什么以下代码片段无法编译通过。编译器报告：
<!-- language: lang-none -->
无法获取 getAString() 的地址
代码如下：
func getAStringPointer() *string {
return &getAString()
}
func getAString() string {
return ""
}
但是，将函数的结果存储在辅助变量中，并返回该变量的地址，编译器就能正常工作。
func getAStringPointer() *string {
var aString = getAString()
return &aString
}
func getAString() string {
return ""
}
英文:
I don't get why the following code snippet isn't compiling. The compiler states:
<!-- language: lang-none -->
cannot take the address of getAString()
<p>
The code:
func getAStringPointer() *string {
return &getAString()
}
func getAString() string {
return ""
}
But, storing the results of the function in auxliary variable and return the address of that variable the compiler behaves OK.
func getAStringPointer() *string {
var aString = getAString()
return &aString
}
func getAString() string {
return ""
}
答案1
得分: 2
你不能将&应用于那样的值（除非它是一个复合字面量）。根据规范：
对于类型为T的操作数x，地址操作
&x生成一个类型为T的指针，指向x。操作数必须是可寻址的*，也就是说，要么是一个变量、指针间接引用或切片索引操作；或者是可寻址结构操作数的字段选择器；或者是可寻址数组的数组索引操作。作为对可寻址要求的例外，x也可以是一个（可能带括号的）复合字面量。如果对x的求值会导致运行时恐慌，那么对&x的求值也会如此。
英文:
You can't apply & to a value like that (unless it's a composite literal). From the spec:
> For an operand x of type T, the address operation &x generates a
> pointer of type *T to x. The operand must be addressable, that is,
> either a variable, pointer indirection, or slice indexing operation;
> or a field selector of an addressable struct operand; or an array
> indexing operation of an addressable array. As an exception to the
> addressability requirement, x may also be a (possibly parenthesized)
> composite literal. If the evaluation of x would cause a run-time
> panic, then the evaluation of &x does too.
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