1 package main
 2
 3 import (
 4  "fmt"
 5 )
 6
 7 type Stu struct {
 8  age int
 9 }
10
11 func (stu *Stu) test() {
12  var i int
13  fmt.Println(i)
14 }
15 func test(stu *Stu) {
16  var i int
17  fmt.Println(i)
18
19 }
20 func main() {
21  var stu Stu
22  stu2 := Stu{1}
23  stu3 := &Stu{1}
24  stu.test()
25  stu2.test()
26  stu3.test()
27  test(&stu)
28  test(&stu2)
29  test(stu3)
30  fmt.Println(stu, stu2, stu3)
31
32 }
33

GNU gdb (GDB) 7.7
Copyright (C) 2014 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
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and "show warranty" for details.
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Type "show configuration" for configuration details.
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For help, type "help".
Type "apropos word" to search for commands related to "word"...
Reading symbols from c...done.
Loading Go Runtime support.
(gdb) l
11      func (stu *Stu) test() {
12              var i int
13              fmt.Println(i)
14      }
15      func test(stu *Stu) {
16              var i int
17              fmt.Println(i)
18
19      }
20      func main() {
(gdb) b 12  (在12行 打上断点
Breakpoint 1 at 0x2019: file /private/tmp/c.go, line 12.
(gdb) b 16 (在16行  打上断点
Breakpoint 2 at 0x20a9: file /private/tmp/c.go, line 16.
(gdb) r   启动调试
Starting program: /private/tmp/c
[New Thread 0x1a0b of process 28719]
[New Thread 0x1b03 of process 28719]
[New Thread 0x1c03 of process 28719]

Breakpoint 1, main.(*Stu).test (stu=0x2210361ed8) at /private/tmp/c.go:12
12              var i int
(gdb) p stu   (启动调试后 在第一个断点处（12行） 停住， 打印stu 的当前地址
$1 = (main.Stu *) 0x2210361ed8
(gdb) c
Continuing.
0

Breakpoint 1, main.(*Stu).test (stu=0x2210361ee0) at /private/tmp/c.go:12
12              var i int
(gdb) p stu  (启动调试后  第2次在12行停住， 打印stu 的当前地址，实际是stu2 变量的地址,stu2,与stu 的地址，很接近， 说明分配在同一处（ 要么都在堆在 要么都在栈上)
$2 = (main.Stu *) 0x2210361ee0
(gdb) c
Continuing.
0

Breakpoint 1, main.(*Stu).test (stu=0x2101f0018) at /private/tmp/c.go:12
12              var i int
(gdb) p stu  (启动调试后  第3次在12行停住， 打印stu 的当前地址，实际是stu3 变
量的地址,stu3的地址,与stu的地址差距很大， 说明不是分配在同一处，在main 函数中，
定义stu3时使用过& 操作符， 所以可以基本肯定 stu3 是分配在堆上 ，而 stu stu2是分配在栈
上
我还有一个疑问 ， 就是 （是不是所有&取地址都会导致变量在椎上分配?），比如在27 28 行
的调用，也对stu ,stu2进行了取址址操作，
下面继续跟踪16行处的代码，stu,stu2 的地址，并没有因为对变量取地址操作，而导致与
 stu3 一样都分配在堆上，
 结论     &Stu{1} 定义的变量是在堆上，
         s:=Stu{} 定义的变量在栈上，即使以后&s, 取s 的地址， 也不会导致s 分配在堆上

$3 = (main.Stu *) 0x2101f0018
(gdb) c
Continuing.
0

Breakpoint 2, main.test (stu=0x2210361ed8) at /private/tmp/c.go:16
16              var i int
(gdb) p stu
$4 = (main.Stu *) 0x2210361ed8
(gdb) c
Continuing.
0

Breakpoint 2, main.test (stu=0x2210361ee0) at /private/tmp/c.go:16
16              var i int
(gdb) p stu
$5 = (main.Stu *) 0x2210361ee0
(gdb) c
Continuing.
0
[Switching to Thread 0x1b03 of process 28719]

Breakpoint 2, main.test (stu=0x2101f0018) at /private/tmp/c.go:16
16              var i int
(gdb) p stu
$6 = (main.Stu *) 0x2101f0018
(gdb) c
Continuing.
0
{0} {1} &{1}
[Inferior 1 (process 28719) exited normally]
(gdb) q