当我将变量传递给golang中的私有方法时，它会创建一个新实例吗？
If I have main function:
var a = "foo"
modify(a)
fmt.Println(a)
where
func modify(s string) error {
s = "bar"
}
will the result be "foo" or "bar"?
如果我有主要功能： p>
var a =“ foo“
modify（a）
fmt.Println（a）
code> pre>
其中 p>
func Modify（s 字符串）错误{
s =“ bar”
}
code> pre>
结果将是“ foo” code>或“ bar “ code>？ p>
div>
None. It won't compile because neither 'foo' nor 'bar' is a single character. But let's say you used double quotes instead.
In Golang, arguments are passed by value (they are copied to the new place in memory - stack or heap), and it does not matter whether it is a private or a public method or arbitrary function. The new instance is modified. The result of your example will be "foo".
In order to modify variable lying outside of the function, you have to explicitly pass the pointer pointing to such variable.
func modify(s *string) {
*s = "bar"
}
...
var a = "foo"
modify(&a)
println(a) // will print "bar"
In this case pointer itself is passed by values (it is copied) but its value (the address of a) still points to the same variable. So the a can by modified through the pointer.