指针解引用在Go中如何工作? [英] How does pointer dereferencing work in Go?
问题描述
我正在通过 http://tour.golang.org/浏览golang教程,以及在示例29
I'm going through the golang tutorials at http://tour.golang.org/, and was experimenting a bit with some things in example 29
供您参考,原始示例复制在这里:
For your reference, the original example is copied here:
package main
import "fmt"
type Vertex struct {
X, Y int
}
var (
p = Vertex{1, 2} // has type Vertex
q = &Vertex{1, 2} // has type *Vertex
r = Vertex{X: 1} // Y:0 is implicit
s = Vertex{} // X:0 and Y:0
)
func main() {
fmt.Println(p, q, r, s)
}
这是非常基础的,它显示了如何创建该新结构Vertex
的实例.不过,示例28 显示了通过指向它的指针来操纵顶点,因此我对示例进行了一些修改并为输出感到惊讶.这是修改:
It's pretty basic, showing how to create instances of this fancy new struct, Vertex
. Example 28, though, shows manipulating a vertex via a pointer to it, so I modified the example a bit and was surprised by the output. Here is the modification:
func main() {
t := *q
q.X = 4
u := *q
fmt.Println(p, q, r, s, t, u, t == u)
}
输出:
{1 2} &{4 2} {1 0} {0 0} {1 2} {4 2} false
让我惊讶的是t
不是{4,2},这似乎意味着更改q.X
会更改q
指向的结构实例.来自C/C ++背景,这对我来说似乎是非常奇怪的行为.
The thing that surprised me is that t
is not {4, 2}, which seems to mean that changing q.X
changed the instance of the struct that q
pointed to. Coming from a C/C++ background, this seems like extremely strange behavior to me.
那么,这里到底发生了什么?为什么使用q.X = 4
更改顶点不会传播到t
?
So, what's actually going on here? Why does using q.X = 4
to change the Vertex not propagate to t
?
推荐答案
t := *q
复制q
指向的结构.
如果要观察从t
到q
的变化,请坚持使用指针:
If you want to observe changes to q
through t
, then stick with a pointer:
func main() {
t := q
q.X = 4
u := *q
fmt.Println(p, q, r, s, t, u, *t == u)
}
这会产生您可能正在寻找的输出.
This produces the output you were probably looking for.
{1 2} &{4 2} {1 0} {0 0} &{4 2} {4 2} true
我不确定您觉得什么特别奇怪. C和C ++的行为方式相同.请考虑以下内容:
I'm not sure what seems extremely strange to you. C and C++ behave the same way. Consider the following:
#include <iostream>
struct Vertex
{
int x;
int y;
};
std::ostream& operator<<(std::ostream& out, const Vertex& v)
{
out << "{ " << v.x << ", " << v.y << " }";
return out;
}
int main()
{
Vertex v = Vertex{1, 2};
Vertex* q = &v;
Vertex t = *q;
q->x = 4;
std::cout << "*q: " << *q << "\n";
std::cout << " t: " << t << "\n";
}
此C ++代码的输出显示相同的行为:
The output of this C++ code shows the same behavior:
*q: { 4, 2 }
t: { 1, 2 }
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