C 不应该'；输出不是2吗

这与

main（）

和

func（）

之间的堆栈不同有关吗

据我所知，指向

'I'

的地址被

'j'

覆盖了，不是吗

---

有人能纠正我的理解吗？

在C中，参数是作为副本传递的。因此

p

和

q

是在

func（&i，&j）中传递的原始地址
&i
和
&j
的副本。所以当你这样做的时候：

    int i = 0, j = 1; 
    void func(int *p, int *q) 
    { 
        p = q; //Here the address of p and q are made same. 
      * p = 2; // value @ address pointed by 'p'(which is q) is now set to 2.
    }     
    int main() //Start of Main
    { 
        func(&i, &j); 
        printf("%d %d n", i, j); 
        return 0; 
    }
when address held in p and q are same , i'm not understanding why the output isn't same as well (2 2).
</code>
If i fire up gdb i see the following
(gdb) p &p 
$4 = (int **) 0x7fffffffebc8   <<Actual address of var p                                                                 (gdb) p &q                                                                               
$6 = (int **) 0x7fffffffebc0   << Actual address of var q                                                       (gdb) n                                                                                          17      }                                                                                    
(gdb) p p                                                                                    
$8 = (int *) 0x601040 <j>    << After the func ends its pointing only to 'j'                                                                
(gdb) p q                                                                                    
$9 = (int *) 0x601040 <j>

p = q; //Here the address of p and q are made same. 

p
和
q
现在确实是一样的，都是
j
的地址。所以当你这样做的时候：

    int i = 0, j = 1; 
    void func(int *p, int *q) 
    { 
        p = q; //Here the address of p and q are made same. 
      * p = 2; // value @ address pointed by 'p'(which is q) is now set to 2.
    }     
    int main() //Start of Main
    { 
        func(&i, &j); 
        printf("%d %d n", i, j); 
        return 0; 
    }
when address held in p and q are same , i'm not understanding why the output isn't same as well (2 2).
</code>
If i fire up gdb i see the following
(gdb) p &p 
$4 = (int **) 0x7fffffffebc8   <<Actual address of var p                                                                 (gdb) p &q                                                                               
$6 = (int **) 0x7fffffffebc0   << Actual address of var q                                                       (gdb) n                                                                                          17      }                                                                                    
(gdb) p p                                                                                    
$8 = (int *) 0x601040 <j>    << After the func ends its pointing only to 'j'                                                                
(gdb) p q                                                                                    
$9 = (int *) 0x601040 <j>

p = q; //Here the address of p and q are made same. 

它将
p
指向的内容（即
j
）设置为
2

i
根本不被此修改，因此它仍然是
0
。您将
i
的地址作为参数
p
传递，但是当您将
p
设置为
q
时，您立即放弃了该地址。
最初在全局范围内，i和j值分别为0和1。函数调用i和j的地址（位置）时传递

* p = 2; // value @ address pointed by 'p'(which is q) is now set to 2.

在函数被调用方中，p和q保存i和j的地址。
分配时，
p=q
q被分配给p，这意味着p也拥有j的地址。然后，
*p=2
设置p所指地址的值。p当前指向j。所以打印时j值为2，i值保持不变。在这个场景中，我们没有任何指向i的指针。我们仅通过i修改i值

希望这会有所帮助：）
理解。谢谢你，伙计，请确保你的问题得到了回答。